Question

Here's a question to test your Physics !

The break supplied to a car produces a de-accelerated motion of 6 m/s² in opposite direction to the motion. If car requires 2 sec to stop after applying breaks, calculate the distance traveled by the car during this time.

Answer 1

SOLUTION:-

Given,

A car produces a de-accelerated motion

=) 6m/s²

It's means retardation, so -6m/s²

Car requires time, t= 2sec.

Assum Initial velocity be u

& final velocity be 0 m/s

Case 1:

So, using first equation of motion

=) v= u+at

=) 0=u+(-6)(2)

=) 0= u-12

=) u= 12m/s

Case 2:

Using second equation of motion for travelled distance by a car:

$s = ut + \frac{1}{2} a {t}^{2} \\ \\ = > s = 12(2) + \frac{1}{2} ( - 6)(2 {}^{2} ) \\ \\ = > s = 24 + (\frac{ - 24}{2} ) \\ \\ = > s = 24 - 12 \\ \\ = > s = 12m$

Hence, the Distance travelled by a car is 12m.

Hope it helps ☺️

Answer 2

✴$\underline\bold{Answer:-}$

Given , Time (t) = 2 sec.

Acceleration (a) = -6ms ²

Final velocity (v) = o ms ²

Initial velocity (u) = ?

Distance (s) = ?

Now , using first equation of motion

=> v = u + at

$\bold{or}$ u = v - at

=> 0 - (-6) (2)

12 m /s.

Now , for distance (s) , we use 2nd equation of motion..

$\bold{i.e}$

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

$s = 12 \times 2 + \frac{1}{2} \times - 6 \times 2 \times 2 \\ s = 24 + \frac{1}{2} \times - 24 \\ s = 24 + ( - 12) \\ s = 12m.$

So , $\bold{distance}$ travelled by the car during this time is $\bold{12m}$.