Physics, asked by jasmangill01, 10 months ago

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



100 m

200 m

300 m

400 m

Answers

Answered by singhpalak552
4

Explanation:

to 20 m/s in 5 seconds.

The distance d is from

d = v(avg) • t

where v(avg) is the average velocity during the time, or

v(avg) = v(initial) + ∆v/2.

Here v(initial) is 0 and ∆v is 20 m/s, so

v(avg) = 20 m/s / 2 = 10 m/s.

d = 10 m/s • 5 s = 50 m.

Answered by Jasleen0599
3

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.

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