A biker reduces the speed from $72 k m h r^{- 1}$ to $36 k m h r^{- 1}$ in $6 s$ . Calculate the acceleration in this case.

1.67 m s^{- 2}

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

- 1.67 m s^{- 2}

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

1.76 m s^{- 2}

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

- 2.67 m s^{- 2}

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

Open in App

Solution

The correct option is B $- 1.67 m s^{- 2}$
Given: initial velocity $u = 72 k m h r^{- 1} = \frac{72 \times 5}{18} = 20 m s^{- 1}$ ,
final velocity $v = 36 k m h r^{- 1} = \frac{36 \times 5}{18} = 10 m s^{- 1}$
time $t = 6 s$
Using v = u + at
Or a = $\frac{10 - 20}{6} = \frac{- 10}{6} = - 1.67 m s^{- 2}$

Suggest Corrections

Similar questions

Q. Starting from rest, a car acquires a speed of

90 k m h^{- 1}

10 s

, calculate the acceleration in this case.

Q. A motorist accelerates from rest to a velocity of

90 k m h r^{- 1}

15

seconds. What is the average acceleration of the car?

Q. A biker reduces the speed from

72 k m h r^{- 1}

36 k m h r^{- 1}

6 s

. Calculate the acceleration in this case.

A wooden block of mass $1 k g$ travelling in a straight line with a velocity of $10 m s^{- 1}$ collides with and sticks to a stationary concrete block of mass $5 k g$ . After the collision they move together in the same straight line. Find the velocity of the combined object.

Q. Starting from rest a car acquires a speed of 90 km/hr in 10 s, calculate the acceleration in this case.

Join BYJU'S Learning Program

A biker reduces the speed from 72 kmhr−1 to 36 kmhr−1 in 6 s. Calculate the acceleration in this case.

The correct option is B −1.67 ms−2Given: initial velocity u=72 kmhr−1=72×518=20 ms−1, final velocity v=36 kmhr−1=36×518=10 ms−1 time t=6 s Using v = u + at Or a = 10−206=−106=−1.67 ms−2

A biker reduces the speed from $72 k m h r^{- 1}$ to $36 k m h r^{- 1}$ in $6 s$ . Calculate the acceleration in this case.