Question

a car moving at 36 km/hr accelerates and covers a distance of 24m if the final velocity of the car is 54 km/hr then calculate acceleration of the car and the time for which the car was accelerating

Answer 1

HEYA!!
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Given that

Initial velocity = 36 km/hr.

= 36× 5/18= 10m/s.

Final velocity = 54 km/hr.

= 54× 5/18

= 15 m/s.

s= 24 m.

Now to find ,

A : ?

T:?

Using thirds equation of motion

$v {}^{2} - u {}^{2} = 2as \\ \\ 15 {}^{2} - 10 {}^{2} = 2 \times 24 \times a \\ \\ 225 - 100 = 48a \\ \\ 115 = 48a$
115/48 = a

Acceleration = 2.39 m/s2.

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Time = ?

Using first equAtion of motion,

V = U + AT

15 =10 + 2.39T

5 = 2.39 T

= 2.09 secs

Approximately 2 seconds


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Answer 2

Answer:

Given that

Initial velocity = 36 km/hr.

= 36× 5/18= 10m/s.

Final velocity = 54 km/hr.

= 54× 5/18

= 15 m/s.

s= 24 m.

Now to find ,

A : ?

T:?

Using thirds equation of motion

115/48 = a

Acceleration = 2.39 m/s2.

----------------------------------------------------------------------------------------------------

Time = ?

Using first equAtion of motion,

V = U + AT

15 =10 + 2.39T

5 = 2.39 T

= 2.09 secs

Approximately 2 seconds