Question

a car is moving with velocity 144km/h comes to rest in 10sec find the retardation and distance traveled to the rest​

Answer 1

Given:-

• Initial velocity,u = 144×5/18 = 40m/s

• Final velocity ,v = 0m/s

• Time taken ,t = 10s

To Find:-

• Distance travelled,s

Solution:-

Firstly we calculate the acceleration of the car

• a = v-u/t

Substitute the value we get

→ a = 0-144/10

→ a = -14.4 m/s²

Here, Negative sign show retardation

Therefore, the acceleration of car is 14.4m/s².

Now, Using 3rd equation of motion

• v² = u² +2as

Substitute the value we get

→ 0² = 40² + 2×14.4 ×s

→ 0 = 1600 + 28.8 ×s

→ s = 1600/28.8

→ s = 55.55 m

Therefore, the distance covered by the car is 55.55 Meters.

Answer 2

Solution

Given,

• initial velocity (u) = 144×5/12 = 40 m /s .
• Final velocity (V) = 0m/s .
• Time taken (T) = 10s .

To find ,

• we have to find here the distance travelled by the car .(s)

How to find ?

• we can find it by first finding its acceleration and then by using the 3rd equation if motion .

So to find the acceleration :-

$= > \red{a = \frac{v - u}{t} }$

Now,

putting the value into equation we get ;

$= > \red{a = \frac{0 - 144}{10} }$

$= > \red{a = - 14.4m \ s}$

so the acceleration is 14.4 m/s .

• here are the acceleration is in negative sign so it shows retradiction. so it will be +14.4 or -14.4

Now to find the distance ,

• we have to use here the third equation of the motion

$\: \: \: \: \: \: \green{ {v}^{2} = {u}^{2} + 2as}$

now ,

putting the value in the equation we get ;

$\red {{0}^{2} = {40}^{2} + 2 \times 14.4 \times s}$

$\red{ = > 0 = 1600 + 28.8 \times s}$

$\red{ = > s = \frac{1600}{28.8} }$

$\red{ = > s = 55.55m}$

The distance covered by the car is 55.55m .