You visited us 1 times! Enjoying our articles? Unlock Full Access!

Instantaneous acceleration

The displacem...

Question

The displacement $x$ of a particle along a straight line at time $t$ is given by $x = a_{0} - a_{1} t + a_{2} t^{2}$ . The acceleration of the particle is:

a_{0}

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

a_{1}

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

2 a_{2}

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

a_{2}

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

Open in App

Solution

The correct option is C $2 a_{2}$
As we know that $a = \frac{d^{2} x}{d t^{2}}$
Given, $x = a_{0} - a_{1} t + a_{2} t^{2}$
$\Rightarrow \frac{d x}{d t} = - a_{1} + 2 a_{2} t$
$\Rightarrow a = \frac{d^{2} x}{d t^{2}} = 2 a_{2}$

Suggest Corrections

Similar questions

x

t

x = a_{0} - a_{1} t + a_{2} t^{2}

The displacement x of a particle along a straight line at time t is given by $x = a_{0} + a_{1} t + a_{2} t^{2}$ .The acceleration of the particle is

x

t

x = a_{0} + a_{1} t + a_{2} t^{3}

a

t = 5

x

t

x = a_{0} + a_{1} t + a_{2} t^{2}

(x)

t

x = a_{0} + \frac{a_{1}}{2} t + \frac{a_{2}}{3} t^{2}

Join BYJU'S Learning Program