Question

a car starts from rest and acquires a velocity of 54 km per hour in 2 seconds find distance travelled by car assume motion of car is uniform

Answer 1

Given:-

• Initial velocity ,u = 0m/s
• Final velocity ,v = 54km/hr = 54×5/18 =15m/s
• Time taken ,t = 2 s

To Find:-

• Distance covered by car ,s

Solution:-

We have to calculate the distance covered by the car . Firstly we calculate the acceleration of the car . Using 1st equation of motion .

• v = u + at

where,

• v denote final velocity
• u denote initial velocity
• a denote acceleration
• t denote time taken

Substitute the value we get

$:\implies$ 15 = 0 = a× 2

$:\implies$ 15 = a×2

$:\implies$ 15/2 = a

$:\implies$ 7.5m/s² = a

$:\implies$ a = 7.5m/s²

So , the acceleration of the car is 7.5m/s²

Now, calculating the distance travelled by the car . Using 3rd equation of motion .

• v² = u² + 2as

substitute the value we get

$:\implies$ 15² = 0² + 2×7.5 × s

$:\implies$ 225 = 0 + 15×s

$:\implies$ 225/15 = s

$:\implies$ s = 225/15

$:\implies$ s = 15 m

• Hence, the distance covered by the car is 15 metres.

Answer 2

Given :-

A  car starts from rest and acquires a velocity of 54 km per hour in 2 seconds

To Find :-

Distance travelled

Solution :-

We know that

1 km/h = 5/18 m/s

54 × 5/18 = 15 m/s

a = v - u/t

a = 15 - 0/2

a = 15/2

a = 7.5 m/s²

Now

v² - u² = 2as

(15)² - (0)² = 2(7.5)s

225 - 0 = 15s

225 = 15s

225/15 = s

15 = s