Question

A particle is moving with constant initial velocity
4 ms-1 till t= 1.5s. Then it accelerates at
10 ms-2 from 1.5 s to 3 s. The distance covered
is​

Answer 1

Explanation:

u=4m/s

t= 1.5s

a=10m/s^2

s=ut+1/2at^2

now u can solve from the data

Answer 2

The distance covered is​ $33.25\ m.$.

Explanation:

Given:

A particle is moving with constant initial velocity 4 ms-1 till t= 1.5s.

it accelerates at 10 ms-2 from 1.5 s to 3 s.

To Find:

The distance covered.

Formula Used:

$S=ut+\frac{1}{2} at^{2}$     ------- formula no.01.

Where,

S = Distance

u = Initial Velocity

a = acceleration

t = time.

Solution:

As given,a particle is moving with constant initial velocity 4 ms-1 till t= 1.5s.

Initial velocity u= 4 m/s.

Time t = 1.5 sec.

Acceleration a =0

As given,it accelerates at 10 ms-2 from 1.5 s to 3 s.

Time =3- 1.5 = 1.5 sec.

Acceleration a =10 $m/s^{2}$.

Applying formula no.01.

The distance covered $= 4\times1.5+\frac{1}{2}\times 0\times\ 1.5^{2}+ 4\times1.5+\frac{1}{2}\times 10\times\ 1.5^{2}$

                                     $= 6+0+ 6+11.25$

                                    $=33.25\ m/s^{2}$

Thus,the distance covered is​ $33.25\ m.$

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