Question

The brakes applied to a car
produce an acceleration of 6 m/s in
the opposite direction to the motion. If
the car takes 2 s to stop after the
application of brakes, calculate the
distance it travels during this time.​

Answer 1

$\dag\:\underline{\sf AnsWer :}$

In the question we are given that, a car produce an acceleration of 6 m/s² in opposite direction to the motion i.e Retardation (a) = -6 m/s². We are also given that,the car takes 2 s to stop after the application of brakes i.e time taken (t) = 2 second. Here final velocity (v) will be 0 beacuse after applying the brake car will stop.

• And we are asked to find the distance travelled by the car but before that we need to find the initial velocity (u) of the car. To calculate the initial velocity we will use the given below formula :

$:\implies \sf Acceleration = \dfrac{Final \: Velocity - Initial \: Velocity}{Time}$

• Now, we have to simply plug in the known values in above formula so that we can calculate the initial velocity of the car :

$:\implies \sf - 6 = \dfrac{0 - u}{2}$

$:\implies \sf - 6 \times 2= 0 - u$

$:\implies \sf -12= 0 - u$

$:\implies \sf -12= - u$

$:\implies \underline{ \boxed{\sf u = 12 \: m/s}}$

• Now to find the distance travelled by the car we shall use the second kinematical equation of motion.

$\dashrightarrow\:\:\sf s = ut + {}^{1} \! / {}_{2} \: at^2$

$\dashrightarrow\:\:\sf s = 12 \times 2 + {}^{1} \! / {}_{2} \: - 6 \times {(2)}^{2}$

$\dashrightarrow\:\:\sf s = 24 \: - 3 \times 4$

$\dashrightarrow\:\:\sf s = 24 \: -12$

$\dashrightarrow\:\: \underline{ \boxed{\sf s = 12 \: m}}$

Answer 2

$\huge\bf{\red{\underline{\underline{\mathcal{AnSwer}}}}}$

Given:-

• Retardation = - 6m/s² ( since, the acceleration is given in an opposite direction, it will be considered retardation)
• time taken = 2 seconds
• final velocity (v) = 0 m/s ( the car is being stopped after the application of breaks)

To find:-

• Distance travelled by this car

Formula to be used:-

$\underline{\boxed{\sf a = \dfrac{v-u}{t}}}$

$\underline{\boxed{\sf s = ut + \dfrac{1}{2}at²}}$

Solution:-

$\dashrightarrow$ -6 = $\sf\dfrac{0-u}{2}$

$\dashrightarrow$ -6×2 = -u

$\dashrightarrow$ u = 12 m/s

Now, Distance:-

$\dashrightarrow$ $\sf S$ = $\sf 12×2$ + $\sf\dfrac{1}{2}$ × (-6)(2)²

$\dashrightarrow$ $\sf S$ = $\sf 24 + (-12)$

$\dashrightarrow$ $\sf S = 12 m$