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, $a_3=10$
Area under the graph represents the change in velocity so
$v_{max}-v_0=at_0+ at_0+ \frac {1}{2}(at_0)+ \frac {1}{2}(2a \times 2t_0)$
$v_{max}-v_0=2at_0 + \frac {1}{2}(at_0)+ 2at_0 = \frac {9}{2}(at_0)$
$v_{max}=v_0+ \frac {9}{2}(at_0)$
$a=mx- \frac {v_{0}^{2}}{x_0}$
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
$m = \frac {v_0^2}{x_0^2}$
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 $
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
$n=2+\sqrt {2}$
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
Answer is (c)
Answer is (b)
$y=2t+t^2-2t^3$
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
$v =2 + \frac {2}{6}- 6(\frac {1}{6})^2 =2 + \frac {1}{3} - \frac {1}{6} =\frac {13}{6}$
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
$= \frac {3v_0}{4}$
It is clear from the option that b and c are correct
$x= \frac {a}{b} ( 1 - e^{-bt})$
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
Acceleration is defined as = $\frac {dv}{dt}$ which is a slope to the velocity –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
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.
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)
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)
Speed =$\frac {dx}{dt}$
Acceleration=$\frac {dv}{dt}$
Distance =$\int vdt$
Hence the answer is (d)
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)
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)
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)
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
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
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}
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 =$u+v$
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 = $u+v-w$
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 = $-u+v=v-u$
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 = $v-u-w$
Velocity of train w.r.t to driver=Velocity of train w.r.t earth -Velocity of driver w.r.t earth
So =0
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
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
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
\[s=x_2-x_1 \\
=\frac{1}{2}g\left( t^2+\frac{1}{m^2}+2\frac{t}{n} \right) -\frac{1}{2}gt^2 \\
=\frac{g}{2m^2}+\frac{gt}{m}\]
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
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$.
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).