A car traveling at a constant speed of 95 km/he is passed by motorcycle travelin

Question

A car traveling at a constant speed of 95 km/he is passed by motorcycle traveling at 140 km/h. Just as the motorcycle is alongside the car, the rider sneers at the car. Exactly 1.0s after the speeder passes, the car accelerates to catch the motorcycle. If the car's acceleration is 2.00m/s2, how much time is needed for the car to overtake the speeder?

I found the t of each and then set them equal to each other but then when I go see how fast the car is going when it catches up the velocity is less than 140km/h so it can't be right! Can you please write out every single?

Explanation / Answer

speed u of car = 95 km/hr = 26.389 m/s
Speed of bike v = 140 km/hr = 38.889 m/s
in 1 second , the bike moves ahead by a distance d = (38.889-26.389) m = 12.5 metres

Now,
Initial speed of car u = 26.389 m/s
acceleration = 2 m/s^2
equating the time , we take,
s = ut+ 0.5 at^2
(12.5+d) = 26.389t + 0.5 x 2 x t^2
12.5+ d = 26.389t + t^2 ---------------eqaution 1

now, bike covers in time t , distance d = 38.889 t
now,
12.5+ 38.889t = 26.389t + t^2

t^2 -12.5 t -12.5 = 0
time taken t = 13.431 seconds

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