Question

Find acceleration of the object if velocity is given by v = (5t2 + 3t) m/s after t = 5seconds.  ​

Answer 1

Relation of velocity w.r.t time:

v = 5t² + 3t

Rate of change of velocity is known as acceleration:

$\sf a = \dfrac{dv}{dt} \\ \\ \sf a = \dfrac{d}{dt}(5t^2 + 3t) \\ \\ \sf a = 10t + 3$

Acceleration of object at t = 5s:

a = 10 × 5 + 3

a = 50 + 3

a = 53 m/s²

Answer 2

Answer:

The acceleration is 53 $m/s^{2}$

Explanation:

Here we have been given to find the acceleration of the given object at t = 5 seconds that has a velocity as follows:

v = $5t^{2} + 3t$

Now we know that we can find the acceleration as below;

a = $\frac{dv}{dt}$

⇒ a = $\frac{d}{dt}$ $(5t^{2} + 3t)$

⇒ a = 5 × 2t + 3

⇒ a = 10t + 3

Putting t = 5 seconds

⇒ a = 10 × 5 +3

⇒ a = 50 +3

⇒ a = 53 $m/s^{2}$

Therefore the acceleration of the object is found to be 53 $m/s^{2}$