The displacement of the particle is given by $x = 3 t^{3} - 2 t^{2} + 5$ in $m$ . The acceleration of the particle at $4 sec$ is

58 m / s^{2}

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

62 m / s^{2}

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

68 m / s^{2}

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

70 m / s^{2}

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

Open in App

Solution

The correct option is C $68 m / s^{2}$
As we know that the acceleration of the particle is given by
$a = \frac{d v}{d t}$ and also, $v = \frac{d x}{d t}$
Given, $x = 3 t^{3} - 2 t^{2} + 5$
So, $v = \frac{d x}{d t} = \frac{d (3 t^{3} - 2 t^{2} + 5)}{d t}$
$\Rightarrow v = 9 t^{2} - 4 t$
also, $a = \frac{d v}{d t} = \frac{d (9 t^{2} - 4 t)}{d t} = 18 t - 4$
Thus, the acceleration of the particle is $18 \times 4 - 4 = 68 m / s^{2}$

Suggest Corrections

Similar questions

x = 3 t^{3} - 2 t^{2} + 5

m

4 sec

x = t + 2 t^{2} + 3 t^{3} - 4 t^{4}

x = 3 t^{3} - 2 t^{2} + 5

m

4 sec

x = 3 t^{3} + 6 t^{2} + 2 t + 5

m

4 sec

(x = 2 t^{2} + t + 5)

Join BYJU'S Learning Program