Chapter 17 : Linear Motion „„„„„ 365

Example 17.6. A burglar’s car had a start with an acceleration of 2 m/s^2. A police vigilant
party came after 5 seconds and continued to chase the burglar’s car with a uniform velocity of 20 m/s.
Find the time taken, in which the police van will overtake the burglar’s car.

Solution. Given : Acceleration of burglar’s car = 2 m/s^2 and uniform velocity of the police
party = 20 m/s.
Let t= Time taken by the police party, to overtake the burglar’s car
after reaching the spot.
First of all, consider the motion of the burglar’s car. In this case, initial velocity (u) = 0 (be-
cause it starts from rest) ; acceleration = 2 m/s^2 and time taken by burglar’s car to travel the
distance of s m (t 1 ) = (t + 5) second.

We know that the distance travelled by the burglar’s car,

(^11222) 02(5)(5)
22
sut at=+ =+× + =+t t ...(i)
Now consider the motion of the police party. In this case, uniform velocity, v = 20 m/s
∴ Distance travelled by the police party,
s = Velocity × Time = 20 × t ...(ii)
For the police party to overtake the burglar’s car, the two distances (i) and (ii) should be equal.
Therefore equating equations (i) and (ii)
(t + 5)^2 = 20 × t
or t^2 + 25 + 10 t = 20 t
or t^2 + 25 – 10 t = 0
or (t – 5)^2 = 0
(t – 5) = 0 ...(Taking square root)
or t= 5 s Ans.
Example 17.7. A train is uniformly accelerated and passes successive kilometre stones
with velocities of 18 km.p.h. and 36 km.p.h. respectively. Calculate the velocity, when it passes the
third kilometre stone. Also find the time taken for each of these two intervals of one kilometre.
Solution. First of all, consider the motion of the train between the first and second kilometre
stones. In this case, distance (s) = 1 km = 1000 m ; initial velocity (u) = 18 km.p.h. = 5 m/s ; and final
velocity (v) = 36 km.p.h. = 10 m/s
Velocity with which the train passes the third km stone
Let v= Velocity with which the train passes the third km, and
a= Uniform acceleration.
We know that v^2 =u^2 + 2as
(10)^2 = (5)^2 + (2a × 1000) = 25 + 2000 a
∴
100 25 (^75) 0.0375 m/s 2
2000 2000
a
−

Now consider the motion of the train between the second and third kilometre stones. In this
case, distance (s) = 1 km = 1000 m and initial velocity (u) = 36 km.p.h. = 10 m/s.
We know that v^2 = u^2 + 2as = (10)^2 + (2 × 0.0375 × 1000) = 175
∴ v= 13.2 m/s = 47.5 km.p.h. Ans.