Physics, asked by Pralabh1396, 1 year ago

A motorcycle and a car start from rest at the same place at the same time and they travel in the same direction. The cycle accelerates uniformly at 1m/sq.m up to a speed of 36kph and car at 0.5 m/sq.s upto a speed of 54kph. Calculate the time and the distance at which the car overtakes the cycle.

Answers

Answered by Fatimakincsem
12

Car will overtake bike after 35 s

Explanation:

The distance from starting point, where bike reaches it final speed 10 m/s is given by  

V^2 = 2 × a × S

Hence S = V^2/(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 15 m/s is given by  

 V^2 = 2 × a × S

Hence S = v^2/(2a) = 225/(2 × 0.5) = 225 m ...............(1)

Time t2 for the car to reach final velocity = v/a = 15/0.5 = 30 s.

Distance travelled by bike = Initial 50 m in 10 s + distance travelled in 20 s with uniform speed 10 m/s

 = 50 + 2× 100 = 250 m ............(2)

From (1) and (2), we know that after 30 s the distance between car and bike = 25 m.

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.

Thus Car will overtake bike after 35 s.

Also learn more

A car travelling at 20 km/h, it speeds up to 60 km/h in 6 seconds. What is its acceleration?

https://brainly.in/question/186505

Answered by mallusoumyapatil
0

Answer:

answer is 35 sec

Explanation:

hope it is helpful

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