Question

The brakes applied to a car produce an acceleration of 8 m/s2

in the opposite direction to

the motion. If the car takes 2s to stop after the application of breaks, calculate the

distance traveled during this time.​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Distance=16\:m}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Acceleration(a) = - 8 \: m/ {s}^{2} \\ \\ \tt: \implies Time(t) = 2 \: sec \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Distance \: travel (s)= ?$

• According to given question :

$\tt \circ \: Final \: velocity = 0 \: m/s \\ \\ \bold{As \: we \: know \: that } \\ \tt: \implies v = u + at \\ \\ \tt: \implies 0 = u +( - 8) \times 2 \\ \\ \tt: \implies - u = - 16 \\ \\ \green{\tt: \implies u = 16 \: m/s} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies {v}^{2} = {u}^{2} + 2as \\ \\ \tt: \implies {0}^{2} = {16}^{2} + 2 \times ( - 8) \times s \\ \\ \tt: \implies - 256 = - 16 \times s \\ \\ \tt: \implies s = \frac{ - 256}{ - 16} \\ \\ \green{\tt: \implies s = 16 \: m} \\ \\ \green{\tt \therefore Distance \: travel \: in \: 2 \: sec \: is \: 16 \: m}$

Answer 2

▶ Solution :

Given :

▪ Deceleration of car = 8m/s$^2$

▪ Car takes 2s to stop.

To Find :

▪ Distance travelled during last 2s.

Concept :

→ This question is completely based on concept of stopping distance and stopping time.

→ Formula of stopping distance and stopping time is given by

$\star\bf\:d_s=\dfrac{u^2}{2a}\\ \\ \star\bf\:t_s=\dfrac{u}{a}$

• u denotes intial velocity
• a denotes retardation

Calculation :

$\implies\sf\:d_t=\dfrac{u^2\times a}{2a\times a}\\ \\ \implies\sf\:d_t=\dfrac{{t_s}^2\times a}{2}\\ \\ \implies\sf\:d_s=\dfrac{(2)^2\times 8}{2}\\ \\ \implies\boxed{\bf{\purple{stopping\:distance=16m}}}$