Question

a car accelerates from rest to reach maximum velocity 20m/s at 5m/s² and then retards to rest at 2m/s². what is the time for which the car is in motion ?

Answer 1

Answer:

Firstly

as acceleration is constant , so using equation of motion

V = U + at1

U = 0 as body start from rest

V = at1

t1 = V/a = 20/5 = 4 sec

second part

V = U + at2

V= 0 as body will come to rest

0 = U - at2 ( here - comes because acceleration is in opposite direction of motion )

U = at2

t2 = U/a = 20/2 = 10 sec

Total time in motion = t1 + t2 = 4 + 10 = 14sec

Answer 2

Given : -

A car accelerates from rest to reach maximum velocity 20m/s at 5m/s² and then retards to rest at 2m/s².

Required to find : -

• Time for which the car is in motion ?

Equation used : -

❶ v = u + at

Here,

• v = Final velocity

• u = Initial velocity

• a = acceleration

• t = time taken

Solution : -

A car accelerates from rest to reach maximum velocity 20m/s at 5m/s² and then retards to rest at 2m/s².

we need to find the time for which the car is in motion ?

So,

From the given data we can conclude that ;

1st Case

Acceleration of the car ( a ) = 5 m/s²

Initial velocity of the car ( u ) = 0 m/s

Final velocity of the car ( v ) = 20 m/s²

Now,

Let's find the time taken by the car ?

Using the equation of motion ;

i.e. v = u + at

➜ 20 = 0 + 5 x t

➜ 20 = 0 + 5t

➜ 20 = 5t

➜ 5t = 20

➜ t = 20/5

➜ t = 4 seconds

Hence,

Time taken by the car ( t1 ) = 4 seconds

Case - 2

Retardation of the car ( a ) = - 2 m/s²

Initial velocity of the car ( u ) = 20 m/s

Final velocity of the car ( v ) = 0 m/s

Now,

Let's find the time taken by the car ?

Using the equation of motion ;

i.e. v = u + at

➜ 0 = 20 + - 2 x t

➜ 0 - 20 = - 2 x t

➜ - 20 = - 2 x t

➜ - 20 = - 2t

Taking - ( minus ) common on both sides

➜ - ( 20 ) = - ( 2t )

- ( minus ) get's cancelled on both sides

➜ 20 = 2t

➜ 2t = 20

➜ t = 20/2

➜ t = 10 seconds

Hence,

Time taken by the car ( t2 ) = 10 seconds

Now,

Let's find the time for which the car is in motion

This implies ;

Time for which car is in motion = t1 + t2

Time for which car is in motion = 4 + 10

Time for which car is in motion = 14 seconds