A car accelerates from rest at $5 {m / s}^{2}$ and then retards to rest at $3 {m / s}^{2}$ . The maximum velocity of the car is $30$ $m / s$ . The distance covered by the car is

150 m

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

240 m

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

300 m

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

360 m

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

Open in App

Solution

The correct option is A 240 m
Given maximum Velocity is 30 m/s
Distance travelled during acceleration is given by
$30^{2} - 0 = 2 \times 5 \times S_{1}$
Hence , $S_{1} = 90 m$
Distance travelled during retardation
$0 - 30^{2} = 2 \times - 3 \times S_{2}$
Hence , $S_{2} = 150 m$
$∴$ Total distance travelled is $90 + 150 = 240 m$

Suggest Corrections

Similar questions

Q. A car accelerates with an acceleration of

2 {m/s}^{2}

for

20 s

from rest on a straight path. Then it decelerates with

1 {m/s}^{2}

until its velocity becomes zero. The total distance covered by the car is -

Q. A car starts from rest and acceleration at

4 {m / s}^{2}

for

5 s

. After that it moves with constant velocity for

25 s

and then retards at

2 m / s^{2}

until it comes to rest. The total distance travelled by the car is

Q. A car acceleration from rest at a constant rate 10

m / s^{2}

for some time after which it accelerates at constant rate 5

m / s^{2}

and then comes to rest. Total time elapsed is 15 s. Maximum velocity of the car is
(i) while average velocity of the car during its deceleration is
(ii) Distance traveled by the car at the instant when its velocity is half of its maximum value is (iii) or (iv).

Q. A car, initially at rest, is given a uniform acceleration of a 0.5

m / s^{2}

for 20 s. The car then moves with uniform velocity for 20 s. Brakes are then applied and the car is brought to rest in10 s. Find the total distance covered by the car.

A car starts from rest and accelerates uniformly at the rate of -1m/s¹ for 5sec. It the maintains rhe velocity for 30 s. then the brakes are applied and the car is retarded to rest in 10 s. Find the maximum velocity attained by the car and the total distance travelled by it.

Join BYJU'S Learning Program

A car accelerates from rest at 5m/s2 and then retards to rest at 3m/s2. The maximum velocity of the car is 30 m/s. The distance covered by the car is

The correct option is A 240 mGiven maximum Velocity is 30 m/s Distance travelled during acceleration is given by 302−0=2×5×S1Hence , S1=90mDistance travelled during retardation 0−302=2×−3×S2Hence , S2=150m∴ Total distance travelled is 90+150=240m

A car accelerates from rest at $5 {m / s}^{2}$ and then retards to rest at $3 {m / s}^{2}$ . The maximum velocity of the car is $30$ $m / s$ . The distance covered by the car is