Question

It was once recorded that jaguar left skid marks that were $290m$ in length. Assuming that the jaguar skidded to a stop with a constant acceleration of $- 3.90\,m/{s^2}$, determine the speed of the jaguar before it began to skid.
(A) $47.6\,m/s$
(B) $38.2\,m/s$
(C) $54.6\,m/s$
(D) $57.6\,m/s$

Answer 1

Hint Here, we know that the constant acceleration value and distance length, Constant acceleration, refers to motion where every second the velocity increases by the same amount. According to that before starting, we should measure the speed.
Useful formula:
Constant acceleration formula,
${v^2} - {u^2} = 2as$
Where,
$v$ and $u$ are final and initial velocities,
$a$ is the acceleration,
$s$ is the distance.

Complete step by step answer
Given by,
jaguar left skid marks $s = 290m$,
comes to rest $v = 0$,
Acceleration of jaguar $a = - 3.9m/{s^2}$
Here, $t$ is irrelevant.
We know that traveling with a constant velocity means that you continuously go at the same velocity in the same direction. If you travel with a constant acceleration, your velocity always changes, but every second it changes by a consistent amount.
We are using,
${v^2} - {u^2} = 2as$
The above equation of motion is valid only for constant acceleration. So, for constant acceleration, the magnitude and direction of acceleration must be constant.
Therefore,
Substituting the given value,
We get,
$0 - {u^2} = 2\left( { - 3.9} \right)\left( {290} \right)$
On simplifying,
Here,
$u = \sqrt {2\left( {3.9} \right)\left( {290} \right)}$
Solving the above equation,
We get,
$u = 47.6\,m/s$
Hence,
the speed of the jaguar before it began to skid

Thus, option A is the correct answer.

Note We calculate the acceleration speed here, the acceleration of an object is the rate of change in its velocity that is speed and direction. Therefore, even if its velocity is unchanged. if its trajectory changes an object will accelerate. However, if the velocity of an object is constant, its acceleration would be zero.