Question

ii). A car moving along a straight road with a
speed of 120 km/hr, is brought to rest by
applying brakes. The car covers a distance
of 100 m before it stops. Calculate (i) the
average retardation of the car (ii) time
taken by the car to come to rest.
[Ans: 50/9 m/sec, 6 sec)​

Answer 1

Given :-

• A car moving along a straight road with a speed of 120 km/hr, is brought to rest by applying brakes. The car covers a distance of 100 m before it stops.

To find :-

• The average retardation of the car
• Time taken by the car to come to rest.

Solution :-

• Initial velocity (u) = 120km/h

• Final velocity (v) = 0

• Distance covered (s) = 100m

→ u = 120 × 5/18

→ u = 100/3 m/s

• According to third equation of motion

→ v² = u² + 2as

→ (0)² = (100/3)² + 2 × a × 100

→ 0 = 10000/9 + 200a

→ - 10000/9 = 200a

→ a = -10000/9 × 200

→ a = - 100/18

→ a = -50/9

Hence,

• Retardation of a car is 50/9 m/s

First equation of motion

→ v = u + at

→ 0 = 100/3 + (-50)/9 × t

→ 0 = 100/3 - 50t/9

→ 50t/9 = 100/3

→ t = 9 × 100/50 × 3

→ t = 3 × 2

→ t = 6 sec

Therefore,

• Time taken = 6 sec

Answer 2

Question:-

A car moving along a straight road with a speed of 120 km/hr, is brought to rest by applying brakes. The car covers a distance of 100 m before it stops. Calculate (i) the average retardation of the car (ii) time taken by the car to come to rest.

Solution:-

Given,

• Car has a velocity of 120 km/h and came to rest.[ v = 0 ]
• It covers a distance of 100m before it stops.

Then,

We have to find the retardation of the car and time taken by the car to come to rest.

Convert kmph to m/s:-

120 kmph = 120 × 5/18

120 kmph = 33.333 m/s (or) 100/3

Now,

u = 100/3 m/s ; v = 0 ; s = 100m ; a = ? ; t = ?

We have to use 2nd equation to find the acceleration.

➣ v² - u² = 2as

➣ 0² - ( 100/3)² = 2(a)(100)

➣ -10000/9 = 200a

➣ a = (200 × 9)/-10000

➣ a = -0.180003744 (or) -50/9m/s²

Now,

We have to use 1st equation of motion to find time.

➣ v = u + at

➣ 0 = 100/3 + (-50)/9(t)

➣ 0 = 100/3 – 50t/9

➣ 50t/9 = 100/3

➣ t = (100 × 9)/(50 × 3)

➣ t = 3 × 2

➣ t = 6sec

∴ time = 6 seconds

acceleration = -50/9m/s²

Verification:-

➣ v = u + at

➣ 0 = 100/3 + (-50/9 × 6)

➣ 0 = 100*3/3*3 - 300/9

➣ 0 = 300/9 - 300/9

➣ 0 = 0