Question

A motor bike running at 90kmh-1
is slowed down to 18kmh-1
in 2.5 s. Calculate
acceleration and distance covered during slow down.

Answer 1

Answer:

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Explanation:

Here motorbike in retardation so

initial velocity (u)= 90 km/h = 25 m/s

final velocity (v) = 18 km/h = 5 m/s

time (t) = 2.5 s

acceleration (a) = ?

distance (s) = ?

a = v - u / t

= 5 - 25 / 2.5

= - 20 / 2.5

a = - 8

for distance apply motions third equation

2 a s = v^2 - u ^2

2 × ( -8 ) × s = ( 5× 5) - ( 25 × 25)

-16 × s = 25 - 625

s = - 600/ - 16

s = 37.5