Question

A car moving with a velocity of 20m/s stopped in a

distance of 40m. If the same car is travelling at double

the velocity, the distance travelled by it for same

retardation is

1) 320 m 2) 1280 m
3) 160 m 4) 640m​

Answer 1

Answer:

160 m

Explanation:

Initial velocity of car is u

v=0

Retardation=a

By v^2-u^2=2as

Or  0−u2=2aS

 S∝u2   (for same a)

Given :  S=40m    u′=2u

∴SS′=u2(u′)2=22

Or  40S′=4

 S′=160m

Answer 2

Answer:

Option 3: 160 m

Explanation:

Given:

• Initial velocity of the car = 20 m/s
• Final velocity of the car = 0 m/s
• Distance travelled = 40 m

To Find:

• Distance travelled by the car if it was travelling at double the velocity for the same retardation

Solution:

First finding the retardation of the car,

By the third equation of motion we know that,

v² - u² = 2as

where v = final velocity

u = initial velocity

a = acceleration/retardation

s = distance travelled

Substitute the data,

0² - 20² = 2 × a × 40

a = -400/80

a = -5 m/s²

Hence the retardation of the car is -5 m/s².

Now if the velocity of the car was double,

Velocity = 2 × 20 = 40 m/s

Using third equation of motion again,

v² - u² = 2as

Substitute the data,

0² - 40² = 2 × -5 × s

s = -1600/-10

s = 160 m

Hence the distance travelled by the car would be 160 m.