Question

A car initially at rest starts moving with a constant acceleration of 0.5 ms⁻¹ and travels a distance of 25 m. Find :
(i) its final velocity
(ii) the time taken

Answer 1

The final velocity attained by the car is 5 ms⁻¹, and the time taken by it is 10 seconds.

• Given, a car is at rest,

=> Initial Velocity (u) = 0

Acceleration (a) = 0.5 ms⁻²

Distance (S) = 25 m

Final Velocity (v) = ?

i) Applying the formula,

v² = u² + 2as, we get

v² = 0 + 2 × 0.5 ms⁻² × 25 m

=> v² = 25 m²s⁻²

=> v = √25 m²s⁻²

=> v = 5 ms⁻¹

ii) Let t be the time taken by the car.

Now, applying the formula,

a = (v - u) / t, we get,

0.5 ms⁻² = ( 5 ms⁻¹ - 0) / t

=> t = 5 ms⁻² / 0.5 ms⁻¹

=> t = 10 s

Answer 2

Answer:

The final velocity is 5 m/s and

The time taken is 10 s.

Explanation:

Given, Initial velocity u = 0,

Acceleration a = 0.5 m/s²

Distance travelled S = 25 m.

(i) From equation of motion v² = u² + 2aS

v² = (0)² + 2 x 0.5 x 25

or, v² = 25

or, Final velocity v =

$\sqrt{25} = 5m {s}^{ - 1}$

(ii) From equation of motion v = u + at

5 = 0 + 0.5 x t

or, 0.5 t = 5

Therefore, time taken

$t \: = \frac{5}{0.5} = 10s$