Question

A car starting from rest attains a velocity of 20m/s in 5 sec find distance convered by the car

Answer 1

Answer:

The distance covered by the car is 50m

Relations to be noted:

v = u + at

$s = ut + \frac{at^{2} }{2}$ where v is the final velocity, u is the initial velocity, a is the acceleration, t is the time taken, and s is the displacement.

Explanation:

Given, final velocity  = 20m/s

           time taken = 5s

           initial velocity = 0

Now acceleration a = v/t = 4 m/$s^{2}$.

Now to obtain the distance coverd we will use the second relation,

                        $s = \frac{4*5^{2} }{2}$ = 50m

Answer 2

Answer:

Concept:

The third equation of motion is used to calculate displacement.

Explanation:

We'll need to utilize the equations of motion for this. We must first determine the acceleration

So the equation goes

$\begin{aligned}&a=v-u / t \\&a=20-0 / 5 \\&a=20 / 5 \\&a=4 \mathrm{~ms}^{\wedge} 2\end{aligned}$( now we got the acceleration).

The third equation of motion is used to calculate displacement.

$\begin{aligned}&s=u^{*} t+1 / 2^{*} a^{*} t^{\wedge} 2 \\&s=0^{*} 5+1 / 2^{*} 4^{*} 5^{\wedge} 2 \\&s=1 / 2^{*} 4^{*} 5^{\wedge} 2 \\&s=2^{*} 5^{\wedge} 2 \\&s=50 \text { meters }\end{aligned}$

Hence the distance covered by the car is 50 meters.

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