Question

A Car is moving in a straight line with a speed of 18 km/h. It is stopped in 5s by applying the
brakes. Find:
(i) The speed of the car in m/s
(ii) The retardation
Salve
(iii) The speed of the car after 2s of applying the brakes.
​

Answer 1

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

$\green{\tt{\therefore{Velocity\:in\:(m/s)=5\:m/s}}}$

$\green{\tt{\therefore{Retardation=1\:m/s^{2}}}}$

$\green{\tt{\therefore{Velocity\:of\:car\:after\:2\:sec=3\:m/s}}}$

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

$\green{\underline \bold{Given :}} \\ \tt: \implies Initial\:velocity(u) = 18 { \: m/s} \\ \\ \tt: \implies Time(t) = 5\: sec \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt:\implies Velocity\:in\:m/s=?\\\\ \tt:\implies Accleration(a)=? \\\\\tt: \implies Velocity\:of\:car\:in\:next\:2\:sec=?$

• According to given question :

$\green{\tt\circ\:Initial\:velocity=18\times \frac{5}{18}=5\:m/s}\\\\ \tt\circ\:Final\:velocity=0\:m/s \\\\\bold{As \: we \: know \: that} \\ \tt: \implies v = u + at \\ \\ \tt: \implies 0 = 5+ a \times 5 \\ \\ \tt: \implies -5=a\times 5\\\\ \tt:\implies a=\frac{-5}{5}\\ \\ \green{\tt: \implies a = -1 \: m/s^{2}}$

$\tt \circ \: Acceleration = -1 { \: m/s}^{2} \\ \\ \tt \circ \: Time = 2\: sec \\ \\ \tt \circ \: Initial \: velocity = 5 \: m/s \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies v = u + at \\ \\ \tt: \implies v = 5 +( -1 )\times 2 \\ \\ \tt: \implies v =5 -2\\ \\ \green{\tt: \implies v =3 \: m/s}$

Answer 2

GIVEN :

A Car is moving in a straight line with a speed of 18 km/h. It is stopped in 5s by applying the brakes.

To FIND :

◾(i) The speed of the car in m/s

◾(ii) The RETARDATION Salve

◾(iii) The speed of the car after 2s of applying the brakes.

SOLUTION :

(i) Given velocity (u) = 18kmh¯¹.

For conversion of speed from kmh¯¹ to ms¯¹,we multiply 5/18.

So, 18 kmh¯¹ ×5/18 = 5 ms¯¹.

(ii) Given initial velocity (u) in ms¯¹ = 5ms¯¹.

As brakes applied, final velocity (v) =0ms¯¹.

Time taken (t) = 5 seconds.

By using Equation of motion i.e v = u+at, we get

v = u + at

0= 5 + a ×5

-5 = a×5

- 1 ms¯² = a

Therefore, acceleration is (-1) ms¯².

(iii) As we know,

u = 5 ms¯

a = - 1 ms¯²

t = 2 sec

Using Equation of motion,

v = u + at

v = 5 + (-1)×2

v = 5 - 2 = 3 ms¯¹

Therefore, speed of the car after 2 seconds of applying brakes is 3 ms¯¹.