Question 1
A particle moves in a straight line according to the relation
$x=t^3-4t^2+3t$
Find the acceleration of the particle at displacement equal to zero.
a. (-8,-2,10)
b. (-1,-2,10)
c. (8,2,10)
d. (1,2,10)
Solution
Given in the question that x=0. Thus
$t^3-4t^2+3t=0$
$t(t-1)(t-3)=0$
so, t=0,1,3
Now acceleration
$x=t^3-4t^2+3t$
$ \frac {dx}{dt} = 3t^2 -8t+3$
$ \frac {d^x}{dt^2} =6t -8$
So,
At t=0, $a_0=-8$
At t=1, $a_1=-2$
At t=3, $a_3=10$
Question 2
A body starts at initial velocity $v_0$ in a straight line with acceleration as shown below in the graph.
Find the maximum velocity reached.
a. $v_0+4at_0$
b. $v_0+ \frac {9}{2}(at_0)$
c. $v_0 - \frac {9}{2}(at_0)$
d. $v_0 - 4at_0$
Solution
Area under the graph represents the change in velocity so
$v_{max}-v_0=at_0+ at_0+ \frac {1}{2}(at_0)+ \frac {1}{2}(2a \times 2t_0)$
$v_{max}-v_0=2at_0 + \frac {1}{2}(at_0)+ 2at_0 = \frac {9}{2}(at_0)$
$v_{max}=v_0+ \frac {9}{2}(at_0)$
Question 3
A particle moves in a straight line with acceleration described by equation given below
$a=mx- \frac {v_{0}^{2}}{x_0}$
If the initial velocity and displacement are $(v_0, 0)$ and at any time $t_0$ velocity and displacement are $(0, x_0)$ the value of constant m is
Solution
$a=mx- \frac {v_{0}^{2}}{x_0}$
Now $a = \frac {dv}{dt} = \frac {dv}{dx} \frac {dx}{dt} = v \frac {dv}{dx}$
$v \frac {dv}{dx}=mx- \frac {v_{0}^{2}}{x_0}$
$vdv = (mx- \frac {v_{0}^{2}}{x_0})dx$
$\int_{v_0}^{0} vdv = \int_{0}^{x_0}(mx- \frac {v_{0}^{2}}{x_0})dx$
$ - \frac {v_0^2}{2} =\frac {mx_0^2}{2} - v_{0}^{2}$
or
$m = \frac {v_0^2}{x_0^2}$
Question 4
A boat moves with the stream of water from point A and B and it return back with the same speed. Velocity of boat relative to water is $\eta$ times the velocity of the water. Velocity of the water is 1m/sec. Find out the average speed in the whole iternary
Solution
When the boat moves from A to B
Speed of the boat= $ \eta +1$
Let $t_1$ be the time taken
Then $ \eta +1= \frac {AB}{ t_1} $
Similarly when the boat moves from point B to A
Speed of the boat= $ \eta -1$
Let $t_2$ be the time taken
Then $ \eta -1= \frac {AB}{t_2}$
Average Velocity
$ =\frac {2AB}{t_1+ t_2}$
Substituting the values from above
Average velocity= $\eta ^2 -1 $
Question 5
A body is freely falling under the action of gravity. It covers half the total distance in the last second of its fall. If it falls for n second, then value of n is
a. $2$
b. $2 + \sqrt {2}$
c. $3$
d. $2 - \sqrt {2}$
Solution
Let h be the total distance travelled in n second then,
$h = \frac {1}{2}gn^2$
In the last second i.e., (n-1)th second distance covered by it is
$h^' = \frac {h}{2} = \frac {1}{4} gn^2$
hence
$ \frac {1}{2}g(n-1)^2 = \frac {1}{4} gn^2$
Solving for n we now get a
Quadratic equation
in n
$n^2-4n+2=0$
Positive root of this equation is
$n=2+\sqrt {2}$
Question 7
Two particle A and B start with the same velocity $v=v_0$ at x=0.They are accelerated per the graph shown above. Which particle has the maximum magnitude of the velocity at $x=x_0$
a. A
b. B
c. A & B will have same velocity
d. None of the above
Solution
Since the graph is a straight line
So for A
$a=-k_1 x$
Where $k_1$ is the constant of the straight line
Now $a = \frac {dv}{dt} = \frac {dv}{dx} \frac {dx}{dt} = v \frac {dv}{dx}$
So
$v \frac {dv}{dx} = -k_1x$
$v dv = ( -k_1 x)dx$
Integrating both sides with in the limit $x=0$ and $x= x_0$
$v^2=v_0^2 - k_1 x_0^2$
Similarly for the particle B
$v^2=v_0^2 - k_2x_0^2$
Now from the graph it is clear that
$k_2 > k_1$
So Particle B has the high Magnitude of the velocity
Question 8
Distance and displacement of a moving object have same magnitude when
a. When object moves in circular motion
b. When object moves along a zig-zag path
c. When object moves along straight line and always moves along the same direction
d. When it moves along straight line but the distance is not always same
Solution
Answer is (c)
Question 9
The greatest possible acceleration or deceleration a train may have is a and its maximum speed is v. Find the maximum time in which the train can get from one station to the next if the total distance is s
Solution
Answer is (b)
Question 10
Displacement(y) of the particle is given by
$y=2t+t^2-2t^3$
the velocity of the particle when acceleration is zero is given by
a. $\frac {5}{2}$
b. $ \frac {9}{4}$
c. $ \frac {13}{6}$
d. $ \frac {17}{8}$
Solution
$y=2t+t^2-2t^3$
Velocity is given
$v=\frac {dy}{dt}=2+2t-6t^2$
$a=\frac {dv}{dt}=2-12t$
Now acceleration is zero
$2-12t=0$
$t=\frac {1}{6}$
Putting this value velocity equation
$v =2 + \frac {2}{6}- 6(\frac {1}{6})^2 =2 + \frac {1}{3} - \frac {1}{6} =\frac {13}{6}$
Question 11
A particle is set in motion at t=0 such that velocity varies as $v= v_0(1 - \frac {t}{2})$ where $v_0$ is a positive constant. Find the distance, displacement covered by the particle in first 3 sec
Solution
Distance is given by
$s = \int_{0}^{3} vdt = \int_{0}^{3} v_0|1 - \frac {t}{2}|dt$
As v here is modulus of velocity
$=\int_{0}^{2} v_0(1 - \frac {t}{2})dt + \int_{2}^{3} v_0(\frac {t}{2} -1)dt = \frac {5v_0}{4} $
Displacement is given by
$r = \int_{0}^{3} vdt = \int_{0}^{3} v_0(1 - \frac {t}{2})dt$
As v here is velocity with positive or negative sign
$= \frac {3v_0}{4}$
Question 12
The velocity, displacement, acceleration of a particle in one dimensional motion is given as
$v_1,x_1,a_1 $ at $t=t_0$
$v_2,x_2,a_2 $ at $t=t_0 + \Delta t$
which of the following is correct
Solution
It is clear from the option that b and c are correct
Question 13
The displacement time equation for a particle in linear motion is given as
which of the following option is correct
a. The velocity and acceleration of the particle at t=0 is $a$ and $-ab$ respectively
b. The velocity will be decreasing as the time increases
c. The displacement of the particle will fall between
d. The maximum acceleration in the motion is $-ab$
Solution
$x= \frac {a}{b} ( 1 - e^{-bt})$
Velocity
$ v=\frac {dx}{dt} =ae^{-bt}$
Acceleration
$w = \frac {dv}{dt} = -abe^{-bt}$
So, at t=0
$v=a$ and $w=-ab$
Also
$v = \frac {a}{e^{bt}}$
So with time ,velocity decreases
$v = -\frac {ab}{e^{bt}}$
Similarly acceleration decreases with time and maximum acceleration is -ab
Now at t=0
$x=0$
when $t -> \infty$
$x= \frac {a}{b}$
So,
$ 0 \leq x \leq \frac {a}{b}$
All the four options are correct
Question 14.
Consider the figure given below
a.Acceleration is positive during A to C and negative during C to B
b.Acceleration is maximum at C
c.Velocity is maximum at C
d. Average acceleration is zero during the journey
Solution
Acceleration is defined as = $\frac {dv}{dt}$ which is a slope to the velocity â€“time diagram
From the graph, it is clear that slope is positive from A to C and negative from C to B
In option .B acceleration is maximum at c is not a correct option because acceleration becomes zero at point c.
This is because zero slope means zero acceleration and the slope is indeed zero at point C.
Now from the graph, it is clear that velocity is maximum at point C
Now Average acceleration is defined =$\frac {change \; in \; velocity}{Time}$
As change in velocity is zero during the journey, so Average acceleration is zero for the journey.
So all options are correct
Question 15.
Which one of the following statement(s) is true?
a.A body moving with uniform speed can have variable velocity
b.A body moving with uniform velocity can have variable speed
c.Average velocity is always equal to instantaneous velocity
d.x-t graph can be a straight line parallel to position axis
Solution
Only option (a) is correct since direction of body moving with constant speed can have directions changing with time and hence it can have variable velocity.
Option b is wrong because magnitude of body moving with uniform velocity remains same and so does its speed remain constant.
Option c is wrong as average velocity and average velocity of a particle along a path are not always same.
Option d is wrong because x-t graph parallel to position axis indicated that the position of the given object is changing at a given instant of time.
Question 16.
Which one of the following statement is correct?
a. A body has constant speed but varying velocity
b. A body has constant speed but varying acceleration
c. A body having constant speed cannot have acceleration
d. None of the above
Solution
In a
uniform circular motion, speed is constants but velocity vector is changing continuously and also acceleration vector is also changing continuously. So all of them are wrong
Hence the correct option is (a)
Question 17.
A body moves along a semicircular track of Radius R. Which of the following statement is true
a. Displacement of the body is $2R$
b. Distance travelled by the body is $\pi R$
c. Displacement of the body is $\pi R$
d. none of the above
Solution
Displacement is measured by the length of line joining the initial and final point.
While distance is the length measured along the path
So in this case
Displacement=$2R$ and distance=$\pi R$
Hence the answer is (a) and (b)
Question 18.
Which of the following is false
a. The speed of the particle at any instant is given by the slope of the displacement-time graph
b. The distance moved by the particle in a time interval from t
_{1}Â to t
_{2}Â is given by the area under the velocity -time graph during that time interval
c. Magnitude of the acceleration of the particle at any instant is given by the slope of the velocity time graph
d. none of the above are true
Solution
Speed =$\frac {dx}{dt}$
Acceleration=$\frac {dv}{dt}$
Distance =$\int vdt$
Hence the answer is (d)
Question 19.
Â A particle is going moving along x-axis. Which of the following statement is false
a. At time t
_{1}Â (dx/dt)
_{t=t1}=0,then (d
^{2}x/dt
^{2})
_{t=t1}=0
b. At time t
_{1}Â (dx/dt)
_{t=t1}Â < 0 then the particle is directed towards origin
c. If the velocity is zero for a time interval, the acceleration is zero at any instant within the time interval.
d. At time t
_{1}Â (d
^{2}x/dt
^{2})
_{t=t1}Â < 0 then the particle is directed towards origin
Solution
If the velocity is zero at any point that does not mean acceleration will become zero at that point
Example when a body is thrown vertically upwards, its velocity becomes zero at highest point but acceleration does not become zero
If the velocity is negative then body is moving towards origin.
If the acceleration is negative that does not mean it is moving towards origin for example when a body is thrown vertically upwards its acceleration is negative but it is moving away from origin
Hence the answer is (b) and (c)
Question 20.
A particle starts at time t=0 from x=0 along the positive x-axis with constant speed v .After time t,it return back towards the origin with the speed 2v and reaches the origin in t/2 sec .Which of the following is true for the whole process
a. Average velocity is zero for the whole process
b. Average speed is $\frac {4v}{3}$ for the whole process
c. Displacement at time t is equal to vt
d. Displacement at time $ \frac {3t}{2}$ is $2vt$
Solution
Since net displacement is zero, average velocity is zero for the whole process
Total distance travelled in forwards direction is vt....total time taken is t
Total distance travelled in backwards direction is vt....total time taken is t/2
So total distance in the journey=2vt
Total time taken=$ \frac {3t}{2}$
$Average \; speed=\frac {total \; distance}{total \; time}$
$=\frac {4vt}{3t}$
$=\frac {4v}{3}$
hence answer is (a),(b) and (c)
Question 21.
A car, starting from rest is accelerated at constant rate a until it attains speed v. It is then retarded at a constant rate b until it comes to rest. which of the following is true
a. the average speed for the whole motion is $ \frac {av}{2b}$
b. the average speed for the whole motion is $ \frac {v}{2}$
c. Total time taken for the journey is $ v(\frac {1}{a}+\frac {1}{b})$
d. none of the above
Solution
the distance s_{1} covered by the car during the time it is accelerated is given by
2as_{1}=v^{2}
Which gives
s_{1}=v^{2}/2a
Similarly in the decelerated motion, distance covered is
2bs_{2}=v^{2}
Which gives
s_{2}=v^{2}/2b
So total distance travelled during the whole journey
s=s_{1}+s_{2}=v^{2}/2(1/a+1/b)
Let t_{1} be the time taken during accelerated motion then
v=at_{1} or t_{1} =v/a
and Let t_{2} be the time taken during decelerated motion then
v=bt_{2} or t_{2}=v/b
Total time taken
t=t_{1}+t_{2} =v(1/a+1/b)
So average speed =total distance/total time taken =s/t=v/2
Hence Answer is (b) and (c)
Paragraph type question
(A) A Man X drops a stone from the fifteen floor of the building .A Man Y ascending in an elevator at a constant speed v=10 m/s passed the Fifteen floor just as the stone is released
Question 22.
Find the position, velocity of the stone as seen by the Man X at time t=2 sec
a. (19m ,19m/s)
b. (19.6m,19.6m/s)
c. (10m,10m/s)
d (11m,12m/s)
Solution
For X, position of stone is given by
$a=g=9.8 m/sec^2$
$x=v_0t + \frac {1}{2} at^2$
or
$x=0 + \frac {1}{2} \times 9.8 \times 4$
=19.6 m
Velocity
$v=v_0+ at$
or v=19.6 m/sec
Question 23
Find the position, velocity of the stone relative to Man Y at 3 sec
a. (39m, 70m/s)
b. (19m, 70m/s)
c. (70m, 39 m/s)
d. (14 m, 29m/s)
Solution
Distance moved by the stone in 3 sec
$x=v_0t + \frac {1}{2}at^2$
or,
x=44 m (downward)
Distance moved by the elevator in 3 sec
=10*3=30 m (upward)
So position w.r.t to Man Y=44+30=70 m/sec
Now Velocity of stone at t=3
=9.8 *3=29m/sec(downward)
Velocity of elevator=10 m/sec(upward)
So velocity of stone w.r.t to Man Y
=29 + 10(since they are opposite direction) =39m/sec
Question 24
Find the acceleration of the stone with respect to Man X and Y.
a. (9.8 m/s
^{2}, 9.8m/s
^{2})
b. (9 m/s
^{2}, 10 m/s
^{2})
c. (10 m/s
^{2}, 10m/s
^{2})
d. None of these
Solution
Man X is at rest. So acceleration will be 9.8 m/s^{2}
Man Y is moving with constant velocity. Again since it is inertial frame, acceleration will be 9.8 m/s^{2}
(C) A gun is mounted on a train roof .Train is travelling with the velocity $v$ in north direction. A car is also moving on a parallel track with train with velocity $w$ in north direction. Two bullet are fired from the muzzle of the gun .Take north as positive and south as negative
Question 25
Bullet one is fired in the north direction with the muzzle velocity $u$. Find the velocity of the bullet as seen from the observer on the earth
a. $u+v$
b. $u-v$
c. $u$
d. $v$
Solution
Velocity of bullet w.r.t to train=$u$
Velocity of train=$v$
Velocity of bullet w.r.t to train=Velocity of bullet w.r.t earth -Velocity of train w.r.t earth
So Velocity of bullet w.r.t earth =Velocity of bullet w.r.t. to train + Velocity of train w.r.t earth =$u+v$
Question 26
Find the velocity of the bullet as seen from the observer on the moving car
a. $u+v-w$
b. $u-v-w$
c. $u$
d. $v$
Solution
Velocity of bullet w.r.t to train=$u$
Velocity of train=$v$
Velocity of car = $w$
Velocity of bullet w.r.t to car = Velocity of bullet w.r.t earth -Velocity of car w.r.t earth = $u+v-w$
Question 27
Bullet one is fired in the south direction with the muzzle velocity u. Find the velocity of the bullet as seen from the observer on the earth
a. $u+v$
b. $v-u$
c. $u$
d. $v$
Solution
Velocity of bullet w.r.t to train=$-u$
Velocity of train=$v$
Velocity of bullet w.r.t to train=Velocity of bullet w.r.t earth -Velocity of train w.r.t earth
So Velocity of bullet w.r.t earth =Velocity of bullet w.r.t to train + Velocity of train w.r.t earth = $-u+v=v-u$
Question 28
Find the velocity of the second bullet as seen from the observer on the moving car
a. u+v-w
b. v-u-w
c. u
d. v
Solution
Velocity of bullet w.r.t to train=$-u$
Velocity of train=$v$
Velocity of bullet w.r.t to car=Velocity of bullet w.r.t earth -Velocity of car w.r.t = $v-u-w$
(D) A train is moving in the west direction with a velocity 15m/s.A monkey runs on the roof of the train against its motion with a velocityÂ 5m/s with respect to train .Take the motion along west as positive
Question 29
Velocity of train relative to its driver
a. 0
b. 15 m/s
c. -15 m/s
d. 20 m/s
Solution
Velocity of train w.r.t to driver=Velocity of train w.r.t earth -Velocity of driver w.r.t earth
So =0
Question 30
What is the velocity of train with respect to monkey
a. 5m/s
b -5 m/s
c. 15 m/s
d -15 m/s
Solution
Given
Velocity of train=15m/s
Velocity of monkey w.r.t train=-5 m/s
Now velocity of monkey w.r.t train=-velocity of train w.r.t monkey
So velocity of train w.r.t monkey=5 m/s
Question 31
Find the velocty of ground with respect to monkey
a. 5 m/s
b. -5 m/s
c. 10 m/s
d. -10 m/s
Solution
Given
Velocity of train=15m/s
Velocity of monkey w.r.t train=5 m/s
Velocity of monkey w.r.t train=velocity of monkey w.r.t ground - velocity of train w.r.t ground
So velocity of monkey w.r.t ground=velocity of monkey w.r.t train + velocity of train w.r.t ground=-5 +15=10m/s
So velocity of ground w.r.t to monkey=-10m/s
Question 32
Water drips from a faucet at a uniform rate of m drops per second. Find the distance x between the two adjacent drops as a function of time t that the trailing drop has been in motion.
a. $x = g\frac {t}{m} + g \frac {1}{2m^2}$
b. $x = g \frac {1}{2m^2}$
c. $x = g \frac {1}{4m^2}$
d. $x=g\frac {t}{2m} + g \frac {1}{2m^2}$
Solution
Answer (a)
Distance x_1 from the faucet of any drop
\(x_1=\frac{1}{2}gt^2\)
m drops in 1 second, so 1 drop takes \(\frac{1}{m}\) second .
Distance \(x_2\) from the faucet of the drop ahead
\(x_2=\frac{1}{2}g\left( t+\frac{1}{m} \right) \) because as it is only 1/m seconds ahead of the previous one.
So, displacement
\[s=x_2-x_1 \\
=\frac{1}{2}g\left( t^2+\frac{1}{m^2}+2\frac{t}{n} \right) -\frac{1}{2}gt^2 \\
=\frac{g}{2m^2}+\frac{gt}{m}\]
Tricky Kinematics Questions
Question 33
A lift is coming from 8th floor and is just about to reach 4th floor. Taking ground floor as origin and positive direction upwards for all quantities, which one of the
following is correct?
(a) x < 0, v < 0, a > 0
(b) x > 0, v < 0, a < 0
(c) x > 0, v < 0, a > 0
(d) x > 0, v > 0, a < 0
Solution
Answer (a)
Displacement-
As the lift is coming in downward direction, displacement will be negative. i.e. x<0
Velocity-
As displacement is negative, velocity must be negative. i.e. v<0
Acceleration-
The lift is about to stop at 4th floor, hence the motion is regarding. i.e. a > 0
Question 34
In one dimensional motion, instantaneous speed v satisfies $0 \leq v < v_0$.
(a) The displacement in time T must always take non-negative values.
(b) The displacement x in time T satisfies $–v_0 T < x < v_0 T$.
(c) The acceleration is always a non-negative number.
(d) The motion has no turning points
Solution
Answer (b)
Here we are given the instantaneous speed v which satisfies $0 \leq v < v_0$. Instananeous speed just tell the magnitude of the instananeous velocity.So instananeous velocity can be negative
Also It means instantaneous speed is increasing or decreasing. So the velocity will be also increasing or decreasing. But here it could happen in any direction.So option (a) ,(b) and (d) may not be correct.
Now maximum displacement can happen with maximum velocity $v_0$ but again it can happen in positive or negative direction.
So maximum displacement in + direction $v_0T$ and maximum displacement in negative direction as $-v_0T$
Hence The displacement x in time T satisfies $–v_0 T < x < v_0 T$.
Question 35
For the one-dimensional motion, described by $x = t–sint$
(a) x (t) > 0 for all t > 0.
(b) v (t) > 0 for all t > 0.
(c) a (t) > 0 for all t > 0.
(d) v (t) lies between 0 and 2
Solution
Here $x = t–sint$
$v = \frac {dx}{dt} =(1 -cos t)$
Also
$a = \frac {dv}{dt}=sint $
Now cos t lies between 1 and -1
For $v_{max}$ at cost minimum i.e., cos t = – 1.
$v_{max} = 1 – (– 1) = 2$
For $v_{min}$ at cos t maximum i.e., cos t = 1
$v_{min} = 1 – 1 = 0$
Hence, v lies between 0 to 2. Verifies the option (d).
$x = t – sin t$
sin t varies between 1 and – 1 for t > 0.
x will be always positive x(t) > 0. Verifies answer (a).
Now as V lies between 0 and 2, So v(t) > 0 is false
Also
a = sin t
sin t varies from – 1 to 1.
So a will varies from – 1 to 1 or a can be (–). So we can discard both (b) and (d)
Hence, verifies option (a) and (d).
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