Question

A light vehicle of mass 50kg starting from rest with an acceleration of 4m/s-2travels for 50m time taken by vehicle is

Answer 1

Answer:

$\boxed{\mathfrak{Time \ taken \ (t) = 5 \ s}}$

Given:

Initial velocity (u) = 0 m/s (As the vehicle starts from rest)

Acceleration (a) = 4 m/s²

Distance travelled (s) = 50 m

To Find:

Time taken (t) by the vehicle

Explanation:

$\sf From \ 2^{nd} \ equation \ of \ motion: \\ \boxed{ \bold{s = ut + \frac{1}{2} a {t}^{2} }}$

Substituting values of s, u & a in the equation we get:

$\sf \implies 50 = 0(t) + \frac{1}{2} \times 4 \times {t}^{2} \\ \\ \sf \implies 50 = \frac{4}{2} {t}^{2} \\ \\ \sf \implies 50 = 2 {t}^{2} \\ \\ \sf \implies 2 {t}^{2} = 50 \\ \\ \sf \implies {t}^{2} = \frac{50}{2} \\ \\ \sf \implies {t}^{2} = 25 \\ \\ \sf \implies {t}^{2} = {5}^{2} \\ \\ \sf \implies t = \sqrt{ {5}^{2} } \\ \\ \sf \implies t = 5 \: s$

$\therefore$

Time taken (t) by the vehicle = 5 seconds

Note:

Mass of the vehicle given in the question does not have any significance.

Answer 2

Answer:

5 sec

Explanation:

Given that, a light vehicle of mass 50kg starting from rest with an acceleration of 4 m/s² travels for 50m

We have to find the time taken by the vehicle.

From above we have u is 0 m/s, a is 4 m/s², s is 50 m and m is 50 kg.

Using the second equation of motion,

s = ut + 1/2 at²

Substitute the values,

→ 50 = 0(t) + 1/2 × 4 × t²

→ 50 = 4t²/2

→ 50 = 2t²

→ t² = 25

→ t² = (5)²

→ t = 5

Hence, the time taken by the vehicle is 5 sec.