Question

A car initially at rest, picksup the velocity of 72 km/hr, over a distance of 25 m. Calculate the acceleration of car and time in which it picks up above velocity. Please don't spam

Answer 1

Given :-

Velocity of car = V = 72 Km/h = 20 m/s

Distance = S = 25 m

Concept :- When a body starts initially at rest then its initial velocity becomes zero and it's moving with final Velocity and while upward motion when a body is thrown upward then it's final velocity becomes zero at the highest point.

Now question,

v² = u² + 2aS

(20)² = 2a(25)

400/50 = a

a = 8 m/s²

Again for time,

v = u + at

20 = 8(t)

t = 20/8

t = 2.5 s

Answer 2

Answer:

Explanation:

Solution,

Here, we have

Initial velocity, u = 0 m/s (As car is at rest)

Final velocity, v = 72 km/h = 72 × 5/18 = 20 m/s

Distance covered, s = 25 m.

To Find,

Acceleration, a = ?

Time taken, t = ?

We know that,

v² - u² = 2as

⇒ (20)² - (0)² = 2 × a × 25

⇒ 400 = 50a

⇒ 400/50 = a

⇒ a = 8 m/s².

Hence, the acceleration of car is 8 m/s².

Now, we know that,

v = u + at

⇒ 20 = 0 + 8 × t

⇒ 20 = 8t

⇒ 20/8 = t

⇒ t = 2.5 seconds

Hence, the time taken by car is 2.5 seconds.