Question

A car running at 72 km/h is slowed down to 18km/h over a distance of 40m. calculate retardation and time for which brakes were applied

Answer 1

$u = 72km {h}^{ - 1} \\ v = 18\: km {h}^{ - 1} \\ s = 40m = .04km \\ {v}^{2} = {u}^{2} - 2as \\ {18}^{2} = {72}^{2} - 2a \times .04 \\ 324 = 5184 - 0.08a \\ 0.08a = 5184 - 324 \\ a = \frac{4860}{0.08} \\ a = 60750 \: km {h}^{ - 2} \\ = \frac{60750 \times 1000}{60 \times 60 \times 60 \times 60} = 4.68 \: m {sec}^{ - 2} \\ \: \\ v = u - at \\ 18 = 72 - 60750t \\ t = \frac{72 - 18}{60750} = \frac{54}{60750} h\\ \frac{54 \times 60 \times 60}{60750} sec= 3.2 \: sec$

Answer 2

Step-by-step explanation:

Here, u = initial velocity = 72 km/hr

v = final velocity = 18 km/hr

s = distance = 40m = 0.04 km

By using the formula v2 = u2 - 2as , we get

v2 = u2 - 2as

=> (18)2 = (72)2 - (2×a×0.04)

=> 324 = 5184 - 0.08a

=> 0.08a = 5184-324

=> 0.08a = 4860

=> a = 4760/0.08 km/hr2

= 60750 km/hr

= 60750× 1000/ 60×60×60×60 m/sec2

= 4.68 m/sec2

By using the formula v = u - at , we get

v = u - at

=> 18 = 72 - 60750t

=> 60750t = 72 - 18

=> 60750t = 54 hr

=> t = 54/60750 hr

= 54×60×60/60750 sec

= 3.2 sec

Hope this solution will help you.