Question

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1)A Car starts from rest and Accelerates
Uniformly over a time of 5.21 Seconds or a Distance of 110 meter . Determine The Acceleration of the Car .

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Answer 1

Distance = s

Acceleration = a

Time = t

Initial velocity = u

GIVEN THAT

s = 110m

a = ?

t = 5.21s

u = 0 m/s

PUTTING SECOND EQUATION OF MOTION :-

s = ut + 1/2 at^2

110 = 0 x 5.21 + 1/2 x a x (5.21)^2

110 = 0 + 1/2 x a x 27.1441

110 = 0 + 1/2 x 27.1441 x a

110 = 0 + 13.572 x a

110 = 13.572 x a

110/13.572=a

8.10 =a

Acceleration = 8.10m/s^2

Abbreviation = Bold x like x this means Multiplication

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Answer 2

Answer:

Acceleration of Car is 8.8 metre/second ²

Explanation:

Given ,

Distance[ S ] = 110 meter

Time [ T ] = 5.21 second

Initial Velocity [ U ] = 0 m/sec

Acceleration [ A ] = ?

Using 2nd Equation of Motion

S = UT + 1/2AT²

110 = 0*5.21 + 1/2*A*(5.21)²

( Round Fig. 5.21 = 5 )

110 = 0*5 + 1/2*A*(5)²

110 = 25A/2

A = 8.8 m/sec²