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

Byju's Answer

Standard XII

Physics

Acceleration

A particle mo...

Question

A particle moves along the $x -$ axis. Its coordinate is given as a function of time by $x = 4 t - 3 t^{2}$ , where $x$ is in metres and $t$ is in second. It's acceleration is:

- 4 m / s^{2}

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

4 m / s^{2}

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

- 3 m / s^{2}

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

- 6 m / s^{2}

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

Open in App

Solution

The correct option is D $- 6 m / s^{2}$
$\begin{matrix} x = 4 t - 3 t^{2} \Rightarrow v = \frac{d x}{d t} = 4 - 6 t \Rightarrow a = \frac{d v}{d t} = - 6 m / s^{2} H e n c e, o p t i o n D i s c o r r e c t a n s w e r . \end{matrix}$

Suggest Corrections

Similar questions

Q. A particle is moving with a velocity

35 m / s

along positive

x

-axis, it's acceleration is towards negative x-axis with the magnitude

4 m / s^{2}

then find the distance covered by the particle in the

9 t h

second.

Q. A particle moves in a circle of radius

4 m

at a speed given by

v = 4 t

, where

v

is in

m/s

and

t

in seconds. The total acceleration of particle at

t = 1 s

is :

Q. For a particle moving along

x

- axis, the acceleration

a

of the particle in terms of its

x

- coordinate is given by

a = - 9 x

, where

x

is in meters and

a

is in

{m/s}^{2}

. Take acceleration, velocity and displacement in positive

x

- direction as positive. The initial velocity of particle at

x = 0

u = + 6 m/s

. The maximum distance of the particle along the positive

x

-direction from origin will be

Q. The position of the particle is given by

x = 3 t^{2} - 6 t + 4

m

. The acceleration of the particle at

10 sec

Q. A particle is moving with a velocity

35 m / s

along positive x-axis.It's acceleration is towards negative X-axis with the magnitude

4 m / s^{2}

then the distance covered by the particle in the

9 t h

second:-

Join BYJU'S Learning Program

A particle moves along the x−axis. Its coordinate is given as a function of time by x=4t−3t2, where x is in metres and t is in second. It's acceleration is:

The correct option is D −6 m/s2x=4t−3t2⇒v=dxdt=4−6t⇒a=dvdt=−6m/s2Hence,optionDiscorrectanswer.

A particle moves along the $x -$ axis. Its coordinate is given as a function of time by $x = 4 t - 3 t^{2}$ , where $x$ is in metres and $t$ is in second. It's acceleration is:

The correct option is D $- 6 m / s^{2}$
$\begin{matrix} x = 4 t - 3 t^{2} \Rightarrow v = \frac{d x}{d t} = 4 - 6 t \Rightarrow a = \frac{d v}{d t} = - 6 m / s^{2} H e n c e, o p t i o n D i s c o r r e c t a n s w e r . \end{matrix}$