Question

1.A car is travelling at a constant speed of 72 kmh−1
and passes a stationary police car. The police car
immediately gives chase, accelerating uniformly to reach a speed of 90 kmh−1
in 10 s and continues at
this speed until he overtakes the other car. Find
(a) the time taken by the police to catch up with the car,
(b) the distance travelled by the police car when this happens.
2. Two trains A and B, starting together from rest, arrive together at rest 10 minutes later. Train A
accelerates uniformly at 0.125 ms−1
for 2 minutes, continues at the steady speed reached for another
4 minutes and the retards uniformly to rest. Train B accelerates uniformly for 5min and then retards
uniformly to rest. Draw both journeys on the same v – t graph and find
(a) the distance (in m) travelled,
(b) the acceleration of train B,
(c) the distance between the two trains after 3 minutes.

Answer 1

The kinematic equation for the first car

$$s_1=v_1t$$

$72km/h=20m/s$

$$s_1=20t$$

The kinematic equation for the police car, accelerating uniformly to reach a speed of 90 kmh−1 

$$v_{pol1}=at_1$$

$$s_{pol1}={at_1^2\over 2}$$

Given

$$v_{pol1}=90km/h=25m/s, t_1=10s$$

Then

$$a={v_{pol1}\over t_1}={25m/s\over 10s}=2.5m/s^2$$

$$s_{pol1}={2.5m/s^2\cdot(10s)^2\over 2}=125m$$

The kinematic equation for the police car with the constant speed

$$s_{pol2}=v_{pol1}t_2$$

The police car overtakes the other car

$$s=s_{pol1}+s_{pol2}$$

$$t=t_1+t_2$$

Substitute

$$20m/s\cdot(10s+t_2)=125m+25m/s\ \cdot t_2$$

$$t_2=15s$$

(a)

$t=10s+15s=25s$

The time taken by the police to catch up with the car, is 25 seconds.

(b)

$$s=20m/s\cdot25s=500m$$

The distance travelled by the police car when this happens is 500 m.

2. Train A

$0\leq t\leq120s$

$a_I=0.125m/s^2$

$v_I=a_I\cdot t=0.125t, v_I(120)=0.125m/s^2\cdot120s=15m/s$

$s_I=\dfrac{a_I\cdot t^2}{2}=0.0625t^2$

$s_I(120)=0.0625m/s^2\cdot(120s)^2=900m$

$120s\leq t\leq 360s$

$a_{II}=0$

$v_{II}=15m/s=const,v_{II}(360s)=15m/s$ ,

$s_{II}=s_I(120)+v_{II}\cdot(t-10s)=900m+15m/s\cdot(t-120s)$

$s_{II}(360)=900m+15m/s\cdot(360s-120s)=4500m$

$360s\leq t\leq600s$

$v_{III}=v_{II}+a_{III}\cdot(t-360s)$

$v{III}(600)=0=15m/s+a_{III}\cdot(600s-360s)$

$a_{III}=\dfrac{-15m/s}{240s}=-0.0625m/s^2$

$$s_{III}=s_{II}(360)+v_{II}(360)\cdot(t-360s)+{a_{III}\cdot(t-360s)^2\over2}=$$

$$=4500m+15m/s\cdot(t-360s)-{0.0625m/s^2\cdot(t-360s)^2\over2}$$

$$s_{III}(600)=6300m$$

Train B

$0\leq t\leq300s$

$u_I=a_I\cdot t$

$u_I(300)=300a_I$

$d_I=\dfrac{a_I\cdot t^2}{2}$

$d_I(300)=\dfrac{a_I\cdot (300s)^2}{2}=45000a_I$

$300\leq t\leq600s$

$u_{II}=u_I(300)+a_{II}\cdot (t-300)$

$u_{II}(600)=0=300a_I+300a_{II}=>a_{II}=-a_I$

$d_{II}=d_I(300)+u_I(300)\cdot (t-300)+\dfrac{a_{II}\cdot (t-300)^2}{2}=$

$=45000a_I+300a_I\cdot(t-300)-\dfrac{a_I\cdot (t-300)^2}{2}$

$d_{II}(600)=45000a_I+90000a_I-45000a_I=s_{III}(600)=6300m$

$a_I=0.07m/s^2$

(a) the distance (in m) travelled

$6300m$

(b) the acceleration of train B

$0.07m/s^2$ at first half of time, $-0.07m/s^2$ at second half of time.

(c) the distance between the two trains after 3 minutes.

Train A

$s_{II}(180)=900m+15m/s\cdot(180s-120s)=1800m$

Train B

$d_I(180)=\dfrac{0.07m/s^2\cdot (180s)^2}{2}=1134m$

$1800m-1134m=666m$

The distance between the two trains after 3 minutes is 666 m.