Question 1.
A bus start at Station A from rest with uniform acceleration 2m/sec^{2}.Bus moves along a straight line
a. Find the distance moved by the bus in 10 sec?
b. At what time, it velocity becomes 20m/sec?
c. How much time it will take to cover a distance of 1.6km
Now first step to attempt such question is to visualize the whole process. Here the bus is moving along a straight line and with uniform acceleration
Now what we have
Initial velocity=0
Acceleration=2m/sec^{2}
Now since it is uniform motion we can use given motion formula's in use
$v=u+at$
$s=ut+ \frac {1}{2}at^2$
$v^2=u^2+2as$
a. Distance (s)=?
time(t)=10 sec
So here the most suitable equation is
$s=ut+ \frac {1}{2}at^2$
Substituting given values
$s=(0)10 + \frac {1}{2} 2(10)^2=100 m$
b. Velocity (v) =20 m/s
time (t)=?
So here the most suitable equation is
$v=u+at$
$20=0+2t$
or t=10 sec
c.distance(s)=1.6km=1600m
t=?
So here the most suitable equation is
$s=ut+ \frac {1}{2}at^2$
$1600=\frac {1}{2}(2)t^2$
or t=40 sec
Question 2.
A object is moving along an straight line. The motion of that object is described by
$x=at+bt^2 + ct^3$
where a,b,c are constants and x is in meters and t is in sec.
a. Find the displacement at t=1 sec
b. Find the velocity at t=0 and t=1 sec
c. Find the acceleration at t=0 and t=1 sec
Now first step to attempt such question is to visualize the whole process. Here the object is moving along a straight line and its motion is described by the given equation
Now we
$x=at+bt^2 + ct^3$
Now since its motion is described by the given equation, following formula will be useful in determining the values
$x=\int vdt$
$v=\int adt$
$v= \frac {dx}{dt}$
$a=\frac {dv}{dt}= \frac {d^2x}{dt^2}$
$a=v \frac {dv}{dx}$
a. x=? t=1sec
Here by substituting t=1 in given equation we get the answer
$x=a(1)+b(1)^2+c(1)^3$
$x=a+b+c$ m
b. v=? t=0,v=? t=1
Here we are having the displacement equation, so first we need to find out the velocity equation
So here the most suitable formula is
$v= \frac {dx}{dt}$
or $v=\frac {d}{dt}(at+bt^2+ct^3)$
or $v=a+2bt+3ct^2$
Substituting t=0 we get
$v=a$ m/s
Substituting t=1 we get
$v=a+2b+3c$ m/s
c. a=? t=0,a=? t=1
Now we are having the velocity equation, we need to first find the acceleration equation.
So here the most suitable formula is
$a=\frac {dv}{dt}$
or $a=\frac {d}{dt}(a+2bt+3ct^2)$
$a=2b+6ct$
Substituting t=0 we get
$a=2b$ m/s^{2}
Substituting t=1 we get
$a=2b+6c$ m/s^{2}
Question 3.
An object is thrown vertically upward with an initial velocity of 40m/s.Two second later another object is thrown upward with the same velocity.
Find out following
a.At what height they meet
b.what is the time when they meet
c.what are the velocities of each object when they meet Solution
Let take the origin at the initial position of the object and upward direction as the positive axis. Let's also take time t as time started when the first object was thrown upward.
Now since it is uniform motion we can use given motion formula's in use
$v=u+at$
$s=ut+ \frac {1}{2)}at^2$
$v^2=u^2+2as$
So from equation of motion
First object
$v_1=v_0 - gt$ and $y_1= v_0t- \frac {1}{2}(gt^2)$ ---(1)
Similarly for the second object. We are writing the equation in same time t variable
$v_2=v_0 - g(t-2)$ and $y_2= v_0(t-2)- \frac {1}{2}g(t-2)^2$ ---(2)
Now when they meet each other, then
$y_1= y_2$
$v_0t- \frac {1}{2}(gt^2) = v_0(t-2)- \frac {1}{2}g(t-2)^2$
Solving it
$t= \frac {v_0}{g}+1=5.09$sec
Substituting this value in any above equation will give the value of height
$H=40 \times 5.09- \frac {1}{2} (9.8 \times 5.09 \times 5.09) =76.6$ m
The respective velocities can be found by using the first set of each equation 1 and 2
For First object
$v_1=v_0 - gt =40-(9.8 \times 5.09)=-9.882$ m/sec
For second object
$v_2=v_0 - g(t-2) =40-9.8(5.09-2)=9.718$ m/sec
Question 4.
A particle moves along the x-axis according to the following equation
$x=pt(1-qt)$ where p and q are constants and p > 0, q>0.
Let take i as the unit vector across x-axis
a. Find out the velocity and acceleration vector for the particle
b. What time it will reach its initial point and what will be the total distance traversed Solution
a. Equation of motion is given by
$x=pt(1-qt)$
Velocity = $\frac {dx}{dt}$
So $v=p(1-2qt)$
Acceleration = $\frac {dv}{dt}$
So $a=-2pqt$
Hence,
Velocity vector= $p(1-2qt) \mathbf{i}$
Acceleration vector=$-2pqt \mathbf{i}$
b. Equation of motion is given by
$x=pt(1-qt)$
at t=0,x=0 and $t= \frac {1}{q}$ ,x=0
So at t=1/q ,it will reach it initial position
Equation of velocity
$v=p(1-2qt)$
From the equation we can see that
For $ t < \frac {1}{2q}$ , $v>0$ so speed=$|v|= p(1-2qt)$
At $t=\frac {1}{2q}$ ,$v=0$
For $t > \frac {1}{2q}$ ,$v < 0$ so speed=$|v|=p(2qt-1)$
Distance traveled from time t=0 to t=1/2q
$s_1=\int_{0}^{\frac {1}{2q}} p(1-2qt)dt = \frac {p}{4q}$
Distance traveled from time t=1/2q to t=1/q
$s_2= \int_{\frac {1}{2q}}^{\frac {1}{q}} p(2qt -1)dt = \frac {p}{4q}$
So total distance traveled=$ \frac {p}{4q} + \frac {p}{4q} =\frac {p}{2q}$
Question 5.
A Balloon rises from rest on the ground vertically upwards with a constant acceleration g/8.An object is dropped from the balloon when it has risen to the height h.Find out the time taken by the object to reach the ground.
Since balloon start's from rest, the velocity of the balloon when it risen to height h is given by
$v^2= u^2+2gH$
$v^2=0 + 2 \times \frac {g}{8} \times h$
$v=\frac {1}{2} \sqrt {gh}$
When the object is dropped, it has the velocity this same velocity in upward direction. Taking upward direction as positive,
$-h= \sqrt {gh} t- \frac {1}{2} gt^2$
Where h is negative as we have taken upward as positive axis , t is the time taken taken to reach the ground and acceleration is -g as gravity is active downwards
Equation can be written in the form
$gt^2-\sqrt {gh} t-2h=0$
Solving this equation and taking positive root
$t=2 \sqrt {\frac {h}{g}}$
Question 6.
A Police Motorcycle is moving on a highway with a speed $v_m$ fires a bullet at at thief motorcycle speeding away in the same direction with a speed $v_t (v_t > v_m)$.If the muzzle speed of the bullet is $v_b(v_b > v_t - v_m)$ find out the following
a. What is the speed of the bullet with respect to the observer sitting on the ground?
b. What speed will the bullet hit the thief
c. What will the speed of the bullet with respect to another police motorcycle moving in the same direction at a speed of v
Speed of the Police Motorcycle w.r.t ground = $v_m$
Speed of the Bullet w.r.t to Motorcycle or Muzzle= $v_b$
The speed of the bullet w.r.t ground=Speed of the bullet w r.t to Motorcycle + speed of the motorcycle w.r.t to ground
So Speed of the bullet w.r.t to ground= $v_b + v_m$
Speed with bullet hit the thief=Speed of bullet w.r.t to ground - Speed of Thief w.r t ground
= $v_b + v_m- v_t$
Speed of the bullet w.r t another motorcycle=Speed of the bullet w.r.t ground - Speed of the Motorcycle
= $v_b + v_m - v$
Question 7.
A nut comes loose from a bolt on the bottom of the elevator as the elevator is moving up the shaft at 3 m/s. The nut strikes the bottom of the shaft in 2 sec
Find out the following
a. How far from bottom of the shaft was the elevator when the nut fell off?
b. How far above the bottom of the shaft was the nut .25 sec after the fell off.?
c. How far above the bottom of the shaft was the elevator when the nut fell on the ground?
d. At what height above the bottom of the shaft, nut has zero velocity after the fell off?
e. what is the total distance traveled by nut in it motion after the fell off?
Given g=9.8 m/s^{2}
a. Here the nut initially has the velocity of the elevator when it fell off. Let take upward direction as the positive direction
Then $v_0=3 $m/s at t=0 and $a=-g=-9.8$ m/s^{2}
Now time to hit the ground t=2
So
$H= v_0t+ \frac {1}{2}gt^2$
$H=3 \times 2 + \frac {1}{2} \times (-9.8) \times 2 \times 2 =-13.6$ m
So the bottom of the shaft was 13.6 m below when the nut fell off from elevator.
So elevator was 13.6 m above the shaft when the nut fell off
b. Now Let calculate the nut displacement at t=.25
$H= v_0t+\frac {1}{2}gt^2$
Substituting the values as same above
H=.44m
Thus the nut was above its starting point. This makes sense if we remember the initial velocity as upward.
So the total height above the bottom will be
=.44+13.6=14 m
c. Now the elevator perspective
v=3m/sec upward
So it will move 6m upward in 2 sec.
So total height above the ground will be
=6+13.6
=19.6 m
d. $v^2=u^2+2ah$
From nut perspective
u=3m/sec
a=-g=-9.8m/s^{2}
at v=0
$u^2=2gH$
or H=3*3/2*9.8
=.45m
So Total height above the bottom of shaft=13.6+.45=14.1m
e. Total distance traveled by nut=.45+.45+13.6=14.5m
Question 8.
The displacement of the body x(in meters) varies with time t (in sec) as
$x=- \frac {2}{3}t^2 +16t+2$
find following
a. what is the velocity at t=0,t=1
b. what is the acceleration at t=0
c. what is the displacement at t=0
d. what will the displacement when it comes to rest
e. How much time it take to come to rest.
Given $x=- \frac {2}{3}t^2 +16t+2$ (1)
Velocity=$ \frac {dx}{dt}=\frac {-4}{3} t+16$
so velocity at t=0 =16
and velocity at t=1 =$\frac {-4}{3} +16=\frac {44}{3}$
Acceleration is given as =$ \frac {d^2x}{dt^2}=- \frac {4}{3}$
so acceleration is time independent and it is constant
Displacement at t=0 can be found by simply substituting the values of t=0 in equation (1)
So, Displacement at(t=0)=2
Now $v=- \frac {4}{3}t+16$
when v=0
then $t= \frac {4}{3}$ sec
Displacement can be found by substituting the value of t=4/3 in equation (1)
$=- \frac {32}{27} + \frac {64}{3} +2$
$= \frac {(-32+576+54)}{27}=\frac {598}{27}$ m
Question 9.
A man runs at a speed at 4 m/s to overtake a standing bus. When he is 6 m behind the door at t=0,the bus moves forward and continues with constant acceleration of 1.2 m/s^{2}
find the following
a. how long does it take for the man to gain the door
b. if in the beginning he is 10m behind the door ,will he running at the same speed ever catch up bus?
At t=0 let man's position be the origin $x_{m0}=0$.The bus door is then at $x_{b0}=6$ m
The equation of motion for the man and the bus
$x_m=x_{m0}+v_{m0}+ \frac {1}{2}a_m t^2$
$x_b=x_{b0}+v_{b0}+\frac {1}{2}a_b t^2$
Now given are
$v_{m0}=4$ m/s , $v_{b0}=0$ ,$a_m=0$ , $a_b=1.2$ m/s^{2}
Thus
$x_m=4t$
$x_b=6+.6t^2$
When the man catches the bus
$x_m=x_b$
so $ 4t=6+.6t^2$
or $3t^2-20t+30=0$
Solving the Quadratic Formula
t= 2.3 and 4.4 s
Note here are two positive solutions .This can be explained as. The first time $t_1=2.3$ s, corresponds to his first reaching the door. This is the real answer. However equation we have solved doesn't know he will stop running and board the bus, the equation have continuing up a running at constant speed. He thus goes on past the bus; but since the bus is accelerating, it eventually builds a larger velocity then the man and will catch with him at 4.4 sec
if the initial position of the bus is 10 m then $x_m=x_b$
gives $3t^2-20t+50=0$ which has only complex roots so no real time exits at which man catches up the bus