Question

A car retards uniformly at 5 m/s^2 to stop after 4 seconds.Find the velocity at which it was travelling and distance covered before it stops.

Answer 1

Explanation:

SO, ACCORDING TO THE QUESTION ITS GIVEN :-

A= 5m/s²

V=0

T=4

v=u+at

o=4+(-5)(4)

u=20m/s.

:. initial velocity is 20m/s.

s=ut+1/2 at²

=20(4)+1/2(-5)(4)²

=80+1/2×(-5)×4×4

=80+(-40)

=80-40

40m.

:. distance is 40 m.

hope it helps ❤

Answer 2

The velocity at which it was traveling and the distance covered before it stops is 40m.

Given:

A car retards uniformly at 5 m/s² to stop after 4 seconds.

To Find:

The velocity at which it was traveling and the distance covered before it stops.

Solution:

To find the velocity at which it was traveling and the distance covered before it stops, we will follow the following steps:

As we know,

The relation between velocity and acceleration is given by:

$v = u + at$

a = -5m/s

v = 0

u =?

t = 4 sec

Now,

In putting values we get,

$0 = u - 5 \times 4$

$u = 20m {s}^{ - 1}$

Now,

The relation between acceleration and distance traveled is :

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

$s = 20 \times 4 - \frac{1}{2} \times (5) {4}^{2}$

$s = 80 - \frac{1}{2} \times 16 \times 5$

$s = 80 - \frac{1}{2} 80 = 40m$

Henceforth, the velocity at which it was traveling and the distance covered before it stops is 40m.

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