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Home / Doubts and Solutions/NEET/Physics

A bike and a car start from rest at the same place at the same time in the same direction. The bike accelerates uniformly at 1m/s^2 upto a speed 10m/s and the car at 0.5 m/s^2 upto speed of 15m/s. At what time the car overtakes the bike? 

Asked by Heraumna176 5th June 2018, 8:07 PM
Answered by Expert
Answer:
The distance from starting point, where bike reaches it final speed 10 m/s is given by 
 
v2 = 2×a×S ; hence S = v2/(2a) = 100/2 = 50 m
 
Time t1 for the bike to reach final velocity = v/a = 10/1 = 10 s. 
 
The distance from starting point, where car reaches it final speed 15m/s is given by 
 
v2 = 2×a×S ; hence S = v2/(2a) = 225/(2×0.5) = 225 m ................................(1)
 
Time t2 for the car to reach final velocity = v/a = 15/0.5 = 30 s.
 
In this 30 s duration of car travel, distance travelled by bike = Initial 50 m in 10 s + distance travelled in 20 s with uniform speed 10 m/s
 
= 50 + 20×100 = 250 m ............................(2)
 
From (1) and (2), we know that after 30 s the distance between car and bike = 25 m.
 
After 30 s , both bike and car are moving with uniform speed. Car speed is 15 m/s and bike speed is 10 m/s.
hence additional time to cover this 25 m gap = 25/(15-10) = 5 s.
 
total time is 30+5 = 35 s.
 
Car will overtake bike after 35 s
 
 
Answered by Expert 6th June 2018, 6:08 PM
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