Question

4 A car is traveling at 15 m/s. If brakes are applied so as to produce uniform acceleration of-5 m/s. calculate the time taken by it to stop?​

Answer 1

Answer:

• 3second is the required answer

Explanation:

Given:-

• Initial velocity ,u = 15m/s
• Final velocity ,v = 0m/s
• Acceleration ,a = -5m/s²

To Find:-

• Time taken to stop it ,t

Solution:-

According to the Question

It is given that car is traveling at 15 m/s. If brakes are applied so as to produce uniform acceleration of-5 m/s². we have to calculate the time taken by car to stop it .So we use here the kinematics equation .

• v = u +at

where

v is the final velocity

a is the acceleration

u is the initial velocity

t is the time taken

Substitute the value we get

→ 0 = 15 + -5×t

→ 0-15 = -5t

→ -15 = -5t

→ 15 = 5t

→ t = 15/5

→ t = 3s

• Hence, the time taken to stop the car is 3 second .

Answer 2

Answer:

Given :-

• A car is traveling at 15 m/s. If brakes are applied so as to produce uniform acceleration of - 5 m/s².

To Find :-

• What is the time taken by it to stop.

Formula Used :-

$\clubsuit$ First Equation Of Motion Formula :

$\mapsto \sf\boxed{\bold{\pink{v =\: u + at}}}$

where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• t = Time taken

Solution :-

Given :

• Initial Velocity = 15 m/s
• Final Velocity = 0 m/s
• Acceleration = - 5 m/s²

According to the question by using the formula we get,

$\longrightarrow \sf 0 =\: 15 + (- 5) \times t$

$\longrightarrow \sf 0 =\: 15 - 5 \times t$

$\longrightarrow \sf 0 - 15 =\: - 5 \times t$

$\longrightarrow \sf {\cancel{-}} 15 =\: {\cancel{-}} 5 \times t$

$\longrightarrow \sf 15 =\: 5 \times t$

$\longrightarrow \sf \dfrac{\cancel{15}}{\cancel{5}} =\: t$

$\longrightarrow \sf \dfrac{3}{1} =\: t$

$\longrightarrow \sf 3 =\: t$

$\longrightarrow \sf\bold{\red{t =\: 3\: seconds}}$

${\small{\bold{\purple{\underline{\therefore\: The\: time\: taken\: by\: it\: to\: stop\: is\: 3\: seconds\: .}}}}}$