Question

A car strats from rest it acuires a speed of 25 ms-1 after 20 s. the distance moved by the car during this time is: 31.25 m 250 m 500 m 5000 m

Answer 1

Answer:

$\boxed{\sf Distance \ moved \ by \ car = 250 \ m}$

Given:

Initial velocity (u) = 0 m/s

Final velocity (v) = 25 m/s

Time taken (t) = 20s

To Find:

Distance moved by car (s).

Explanation:

$\sf From \ 1^{st} \ equation \ of \ motion:$

$\boxed{ \bold{v = u + at}}$

Substituting values of v, u & t in the equation:

$\sf \implies 25 = 0 + a(20)$

$\sf \implies 20a = 25$

$\sf \implies a = \frac{25}{20}$

$\sf \implies a = \frac{5}{4} \: m/s ^{2}$

$\sf From \ 2^{nd} \ equation \ of \ motion:$

$\boxed{ \bold{s= ut + \frac{1}{2} a {t}^{2} }}$

Substituting values of u, a & t in the equation:

$\sf \implies s = 0(20) + \frac{1}{2} \times \frac{5}{4} \times {(20)}^{2}$

$\sf \implies s = 0 + \frac{1}{2} \times \frac{5}{4} \times 400$

$\sf \implies s = \frac{1}{2} \times 5 \times 100$

$\sf \implies s = 5 \times 50$

$\sf \implies s = 250 \: m$