An object having a velocity 4.0 m/s is accelerated at the rate of 1.2 m/s² for 5.0 s. Find the distance travelled during the period of acceleration.

Open in App

Solution

Given:
Velocity, u = 4.0 m/s
Acceleration, a = 1.2 m/s²
Time, t = 5.0 s
Distance travelled:
$s = u t + \frac{1}{2} a t^{2}$
$= 4.0 \times 5 + \frac{1}{2} \times 1.2 \times 5^{2} = 20.0 + \frac{1}{2} \times 1.2 \times 25 = 35 m$

Suggest Corrections

Similar questions

Q. An object having a velocity

4.0 m / s

is accelerated at the rate of

1.2 m / s^{2}

for

5.0 s .

Find the distance travelled during the period of acceleration

Q. An object travels 10 s with uniform acceleration along a straight line path. During this period if the velocity of the object is increased from

{5 m s}^{- 1} t o {25 s}^{- 1}

, then find the distance travelled by the body?

An object with an initial velocity of 12 m/s west experiences a constant acceleration of 4 m/ $s^{2}$ west for 3 second. During this time the object travels a distance of :

Join BYJU'S Learning Program

An object having a velocity 4.0 m/s is accelerated at the rate of 1.2 m/s2 for 5.0 s. Find the distance travelled during the period of acceleration.

Given: Velocity, u = 4.0 m/s Acceleration, a = 1.2 m/s2 Time, t = 5.0 s Distance travelled: s=ut+12at2 =4.0×5+12×1.2×52=20.0+12×1.2×25=35 m

An object having a velocity 4.0 m/s is accelerated at the rate of 1.2 m/s² for 5.0 s. Find the distance travelled during the period of acceleration.

Given:
Velocity, u = 4.0 m/s
Acceleration, a = 1.2 m/s²
Time, t = 5.0 s
Distance travelled:
$s = u t + \frac{1}{2} a t^{2}$
$= 4.0 \times 5 + \frac{1}{2} \times 1.2 \times 5^{2} = 20.0 + \frac{1}{2} \times 1.2 \times 25 = 35 m$