You visited us 3 times! Enjoying our articles? Unlock Full Access!

Differentiation

The position ...

Question

The position of the particle is given by $x (t) = (4 t^{2} - 3 t + 2 t^{3})$ , its acceleration will be

4 t + 2 t^{2}

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

12 t^{2} + 8 t

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

8 t - 3 + 6 t^{2}

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

12 t + 8

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

Open in App

Solution

The correct option is D $12 t + 8$
$a = \frac{d v}{d t} = \frac{d^{2} x}{d t^{2}} = \frac{d^{2} (4 t^{2} - 3 t + 2 t^{3})}{d t^{2}} = 12 t + 8$

Suggest Corrections

Similar questions

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

a = 2 t - 4

m / s^{2}

a = 4 t + 3

(m / s^{2})

t

t = 0, u = 3 m / s, x = 2 m

x

t

x = 2 t^{3} - 12 t^{2} + 6 t + 8

x

t

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

Join BYJU'S Learning Program