Question

1. A car moving with over-speed limit constant speed 31.7 m/s passes a police car at rest. The police car immediately takes off in pursuit, accelerating with 10.1 m/s^2. How long does it take for police overtake the speeder?

2. A car moving with over-speed limit constant speed 31.8 m/s passes a police car at rest. The police car immediately takes off in pursuit, accelerating with 9.6 m/s^2. How far from initial point police car will reach the speeder?

3. Two objects, A and B, start from rest. Objest A starts with acceleration 1.6 m/s^2 and 4.0 seconds later after A, object B starts in the same direction with acceleration 3.4 m/s^2. How long will it take for object B to reach object A from the moment when A started to accelerate?

4. Two objects, A and B, start from rest. Object A starts with acceleration 2.5 m/s^2 and 4.7 seconds later after A, object B starts in the same direction with acceleration 6.1 m/s^2. What will be speed of object A at the moment when it will be reached by object B?

Answer 1

Solution:

Ans 1:

Given Values

Speed of Car = v = 31.7 m/s

acceleration of police car = 10.1 m/s^2

find the time the police will take to overtake speeder ?

To catch the speeding car the distance travelled is the same by both car

for speeding car

v = s / t

s = v x t = 31.7 t eq (1)

for the police car

initial velocity (u) = 0

from second equation of motion

s = ut +1/2 (at²)

s = 0 x t +1/2 ( 10.1 t²)

s = (10.1 /2 ) t² eq(2)

from eq(1) and eq (2)

31.7 t = (10.1/2) t²

t = 63.4 / 10.1 = 6.27 sec

police car takes 6.27 sec to overtake speeder