Question

A car is moving at a rate of 72km/hr. how far does the car move it stops after 4 seonds .. do with all proces .

Answer 1

$\bold{ANSWER}$

u (Initial Velocity) = 72 km/h

u = 72 × 5/18

u = 20 m/s

Also,

v (Final Velocity) = 0 (Because car stops)

t (Time taken) = 4 seconds

Now,

Using 1st Equation of motion,

v = u + at

⇒ 0 = 20 + a × 4

⇒ a = - 5 m/s

Also,

Using 2nd Equation of motion,

⇒ s = ut + 1/2 at²

⇒ s = 20 × 4 + 1/2 × (- 5) × 4²

⇒ s = 80 + 1/2 × (- 5) × 16

⇒ s = 80 + (- 5) × 8

⇒ s = 80 - 40

⇒ s = 40 m.

Therefore,

Distance covered by car = 40 m

Answer 2

Required Answer :

The distance travelled by the car before stopping = 40 m

Solution :

we have,

• Initial velocity = 72 km/h
• Final velocity = 0 m/s
• Time = 4 seconds

Need to find,

• Distance travelled by the car before it stops

Required Solution,

Firstly, let's convert the initial velocity from km/h into m/s.

To convert it into m/s, multiply the value by 5/18.

Initial velocity :

⇒ Initial velocity = 72 × 5/18

⇒ Initial velocity = 4 × 5

⇒ Initial velocity = 20 m/s

First equation of motion :

• v = u + at

⇒ 20 = 0 + a × 4

⇒ 20 = 4a

⇒ 20/4 = a

⇒ 5 = a

Acceleration = 5 m/s²

Second equation of motion :

• s = ut + ½ at²

⇒ s = (0)(4) + ½ × 5 × 4 × 4

⇒ s = 5 × 2 × 4

⇒ s = 40

Therefore,

• The distance travelled by the car before stopping = 40 m