Question

A car having an acceleration, \[1m/{{s}^{2}}\] is moving with an initial velocity 10m/s and it stops after 10 seconds as the driver applies brakes. Find the distance travelled by the car.
A. $30\,m$
B. $40\,m$
C. $50\,m$
D. $60\,m$

Answer 1

Hint:In this question the driver applies brakes and the car will stop after 10 seconds it means the velocity starts decreasing as we know that the rate of change of velocity is the acceleration. If we increase the velocity the acceleration will be positive but if we decrease the velocity the acceleration will be negative. The negative acceleration is called retardation.

Complete step by step solution:
Given values: The acceleration of the car is, a \[=-1m/{{s}^{2}}\]
(negative sign represents the car accelerates because the driver applies brakes)
The initial velocity of the car is, \[S=50\text{ }meters\] \[u\text{ }=\text{ }10m/s\]
The final velocity of the car is, \[v=0m/s\] (After 10 seconds the driver will apply the brakes so the car comes to the rest position)

To find out: The distance travelled by the car in 10 seconds
Apply third equation of motion
\[{{v}^{2}}-{{u}^{2}}=2as\]
Where, $V$ is the final velocity of the car, $U$ is the initial velocity of the car, $a$ is the acceleration of the car,$s$ is the distance travelled by the car
Put all the given values in above equations
\[0-{{10}^{2}}=2(-1)s\]
\[\Rightarrow S=50\text{ }meters\]

Another Method:
Apply second equation of motion
\[S=ut+\dfrac{1}{2}a{{t}^{2}}\]
Put all the given values,
\[
S=(10)10+\dfrac{1}{2}(-1){{(10)}^{2}} \\
\Rightarrow S=100-50\\
\therefore S =50meters \\
\]
The distance travelled by the car in 10 seconds is 50 meters

So, the correct answer is option C.

Note:While solving this most of the students tend to make mistakes while taking the acceleration into account. It should be noted that in this case, the acceleration is negative as it is the case of deceleration. A body can only come to perfect halt if the final velocity becomes zero. The stopping time decreases as the magnitude of deceleration increases. Taking the acceleration in the positive can completely ruin the solution. Please take care of the sign.