Question

A car moving with constant acceleration a= 2 m/s2. crosses the observer with speed v= 20m/s. At what
distance s from the observer was the car behind, t = 6 secprior to this point?
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Answer 1

Answer:

156 m

Explanation:

According to the given question,

a = acceleration = 2 m/s^2

u = initial velocity = 20 m/s

t = time = 6 s

s = displacement

And we are asked to find distance after 6 sec.

Here distance is equal to the displacement as it's a linear motion without change in direction.

We know,

• s = ut+1/2 at^2

Now putting the values we get,

• s = 20*6+ 1/2* 2*36

• s = 120+ 36

• s = 156 m

Thus distance of the car from the observer after 6sec is 156 m when given conditions are applied.

Answer 2

Answer:

Hello Dear User__________

Here is Your Answer...!!

____________________

Explanation:

$Given \ that \ acceration=2 \ m \ sec^{-2}\\\\initial \ velocity=20 \ m \ sec^{-1}\\\\time=6 \ sec\\\\From \ second \ equation \ of \ motion \ we \ have \ formula \ for \ displacement\\\\s=ut+\frac{1}{2}at^{2}\\\\Now \ putting \ value \ here\\\\s=20\times6+\frac{1}{2}\times2\times6\times6 \ m\\\\s=120+36 \ m\\\\s=156 \ m\\\\Hence, \ distance \ of \ the \ car \ from \ the \ observer \ after \ 6 sec \ is \ 156 m$

Hope it is clear to you.