Question

A car travelling at 9 m/s accelerates and
attains
a speed of 27m.s in 5s
colulate the acceleration the distance covered in 5s
​

Answer 1

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A car travelling at 9 m/s accelerates and

attains

a speed of 27m.s in 5s

colulate the acceleration the distance covered in 5s.

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s = ut + ½at^2 …. (1)

v^2 = u^2 + 2as …. (2)

v = u + at …. (3)

s = (u + v)t/2 …. (4)

where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.

In this case, we know u = 9m/s, v = 27m/s and t = 5s, so we use equation(3) to find acceleration, a

v = u + at

27 = 9 + 5t, so a = 3.6m/s^2

The distance traveled is found using equation (4)

s = (u + v)t/2 = (9 + 27)(5)/2 = 90m

In summary, the acceleration, a = 3.6m/s^2 and the distance traveled, s = 90m.