You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard VI

Physics

Kinetic Friction

A vehicle of ...

Question

A vehicle of mass $500 k g$ is moving with a velocity of $15 m s^{- 1}$ . It is brought to rest by a retarding force. Find the distance moved by the vehicle before coming to rest, if the sliding friction between the tyres and the road is $3000 N$ .

18.75 m

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

9.2 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10.5 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

16.8 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $18.75 m$
$A = F / m$
$= 3000 N / 500 k g$
$= 6 m / s^{2}$
Retardation is $6 m / s^{2}$
$S = u^{2} / (2 a)$
$= (15 m / s^{2}) / (2 \times 6 m / s^{2})$
Distance moved by vehicle before coming to rest is $18.75 m$
Hence,
option $(A)$ is correct answer.

Suggest Corrections

Similar questions

Q. A vehicle of mass

1000 k g

is moving with a velocity of

15 m s^{- 1}

. It is brought to rest by applying brakes and locking the wheels. If the sliding friction between the tires and the rod is

6000 N

, distance moved by the vehicle before coming to rest is:

Q. When the engine is switched off a vehicle of mass

^{'} M^{'}

is moving on a rough horizontal road with momentum

P

. If the coefficient of friction between the road and tyres of the vehicle is

μ_{k}

, the distance travelled by the vehicle before it comes to rest is

Q. When a car of mass

1200 k g

is moving with a velocity of

15 m s^{- 1}

on a rough horizontal road. Its engine is switched off. How far does the car travel before it comes to rest if the coefficient of kinetic friction between the road and tyres of the car is

0.5

(g = 10 m s^{- 2})

A 1000 kg vehicle moving with a velocity of 20 m/s is brought to rest in a distance of 50 m. Find the acceleration. Calculate the unbalanced force acting on the vehicle

Q. A

1, 000 k g

vehicle moving with a speed of

20 m s^{- 1}

is brought to rest in a distance of

50 m

. Calculate the unbalanced force acting on the vehicle.

Join BYJU'S Learning Program

A vehicle of mass 500 kg is moving with a velocity of 15ms−1. It is brought to rest by a retarding force. Find the distance moved by the vehicle before coming to rest, if the sliding friction between the tyres and the road is 3000N.

The correct option is A 18.75mA=F/m=3000N/500kg=6m/s2Retardation is 6m/s2S=u2/(2a)=(15m/s2)/(2×6m/s2)Distance moved by vehicle before coming to rest is 18.75mHence,option (A) is correct answer.

A vehicle of mass $500 k g$ is moving with a velocity of $15 m s^{- 1}$ . It is brought to rest by a retarding force. Find the distance moved by the vehicle before coming to rest, if the sliding friction between the tyres and the road is $3000 N$ .

The correct option is A $18.75 m$
$A = F / m$
$= 3000 N / 500 k g$
$= 6 m / s^{2}$
Retardation is $6 m / s^{2}$
$S = u^{2} / (2 a)$
$= (15 m / s^{2}) / (2 \times 6 m / s^{2})$
Distance moved by vehicle before coming to rest is $18.75 m$
Hence,
option $(A)$ is correct answer.