A car moving at a constant velocity, comes to rest upon the application of brakes. The motion of the car is shown with the help of the graph. Find the displacement of the car.

160 m

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

180 m

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

150 m

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

260 m

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

Open in App

Solution

The correct option is B $180 m$
We know that the area under the velocity-time graph gives the displacement.
Thus, Area under the graph
=> Area (OABD) + Area (BDC)
=> ( $20$ x $5$ ) + ( $1 / 2$ )( $20$ x $8$ )
=> ( $100 m$ + $80 m$ )
=> $180 m$

Suggest Corrections

Similar questions

Q. A car is moving with a velocity of

90 km/h

. The application of brakes offers a constant retardation of

5 {m/s}^{2}

. Find the distance travelled by the car before it comes to rest.

Q. A car of mass

2000 K g

moving with a velocity of

72 k m / h

is brought to rest by the application of brakes within a distance of

20 m

. Find the average braking force and the time taken for the car to come to rest.

Q. A car is moving on a straight road with some speed. The driver applies brakes so that car comes to rest with a retardation of

5 m / s^{2}

. The distance traversed by the car in last second of its motion is :

A motor car moving with a uniform speed of $20 m s^{- 1}$ comes to rest on the application of brakes after travelling a distance of 10 m. Its acceleration is

The velocity-time graph of a car moving on a straight road is shown. Find the distance travelled by the car after the brakes are applied :

Join BYJU'S Learning Program

A car moving at a constant velocity, comes to rest upon the application of brakes. The motion of the car is shown with the help of the graph. Find the displacement of the car.

The correct option is B 180 mWe know that the area under the velocity-time graph gives the displacement. Thus, Area under the graph => Area (OABD) + Area (BDC) => (20 x 5) + (1/2)(20 x 8) => (100 m + 80 m) => 180 m

The correct option is B $180 m$
We know that the area under the velocity-time graph gives the displacement.
Thus, Area under the graph
=> Area (OABD) + Area (BDC)
=> ( $20$ x $5$ ) + ( $1 / 2$ )( $20$ x $8$ )
=> ( $100 m$ + $80 m$ )
=> $180 m$