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?
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 (at^{2})
s = 0 x t +1/2 ( 10.1 t^{2})
s = (10.1 /2 ) t^{2} eq(2)
from eq(1) and eq (2)
31.7 t = (10.1/2) t^{2}
t = 63.4 / 10.1 = 6.27 sec
police car takes 6.27 sec to overtake speeder
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