Question

car starts from rest and attains velocity of 20 m/s in 20 sec. The distance travelled in the car during this period is: *

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Answer 1

a=v-u/t

a=20–0/5

=20/5

=4ms^2

We have to use the third equation of motion to find displacement.

S=u*t+1/2*a*t^2

S=0*5+1/2*4*5^2

=1/2*4*5^2

=2*5^2

S=50

So,the answer is 50 meters

Answer 2

Given:

The initial velocity of the car, u = 0

The final velocity of the car, v = 20 m/s

The total time taken by the car, t = 20 sec

To Find:

The distance travelled by the car during this period.

Calculation:

- Using the first equation of motion, we have:

v = u + at

⇒ 20 = 0 + a × 20

⇒ a = 20 / 20

⇒ a = 1 m/s²

- Now using second equation of motion, we get:

s = ut + 1/2 at²

⇒ s = 0 × 20 + 1/2 × 1 × (20)²

⇒ s = 400/2

⇒ s = 200 m

- So, the distance travelled by the car during this period is 200 m.