Question

A car is moving in a straight line with the speed $18\,kmph$ . It is stopped in $5\,s$ by applying the brakes. Find the speed of the car after $2\,s$ of applying the brakes.
A. $1\,m{s^{ - 1}}$
B. $2\,m{s^{ - 1}}$
C. $3\,m{s^{ - 1}}$
D. $4\,m{s^{ - 1}}$

Answer 1

Here we have to apply the first equation of motion to get the answer.
The first equation of motion gives the relation between final velocity, initial velocity, acceleration and time.
We have to put the given values when the car comes to rest and when the brakes are applied in the first equation of motion to get the answer.

Complete step by step answer:
In physics, equations of motion are classified as equations that characterise a physical system’s behaviour in terms of its motion as a function of time. There are three motion equations that can be used for deriving components such as displacement, momentum, time and acceleration.
Given,
Speed $= 18\,km{h^{ - 1}}$
Time, $t = 5\,s$
First we have to convert the unit of speed to the standard unit.
$18\,km{h^{ - 1}} = 18 \times \dfrac{5} {{18}} = 5\,m{s^{ - 1}}$
We know that the first equation of motion is:
$v = u + at$
Since, the car comes to rest,
$v = 0$
$u = 5$
$t = 5$
Putting these values in the equation we get:
$a = - 1\,m{s^{ - 2}}$
Since, the brakes are applied after two second:
$u = 5$
$a = - 1$
$t = 2$
Putting these values in the equation we get:
$v = 3\,m{s^{ - 1}}$

So, the correct answer is “Option C”.

Note:
Here we have to remember to convert the unit to a standard unit otherwise the answer would be wrong. Also we have to check whether the final and initial velocity is zero or not.
The longer the acceleration, the greater the velocity variation. Velocity shift is strictly proportional to the time that continuous acceleration occurs. If velocity increases for a given time by a certain amount, twice the time, it can increase by twice the amount. If an object had already begun at a certain velocity, so the old velocity plus this adjustment would be its current velocity.