Single Choice

For motion of an object along the $$x$$-axis, the velocity $$v$$ depends on the displacement $$x$$ as $$v = 3x^2-2x,$$ then what is the acceleration at $$x=2m.$$

A$$48ms^{-2}$$
B$$80ms^{-2}$$
Correct Answer
C$$18ms^{-2}$$
D$$10ms^{-2}$$

Solution

Given $$v=3x^2-2x;$$ differentiating v, we get
$$\dfrac{dv}{dt}=(6x-2)\dfrac{dx}{dt}=(6x-2)v$$
$$\Rightarrow a=(6x-2)(3x^2-2x)$$ Now put $$x=2m$$
$$\Rightarrow a=(6x-2-2)(3(2)^2-2\times2)=80ms^{-2}$$

SIMILAR QUESTIONS

Kinematics

A particle starts from rest with acceleration $$2m/s^2$$. The acceleration of the particle decreases down to zero uniformly during time-interval of $$4$$ second. The velocity of particle after $$2$$ second is?

Kinematics

When acceleration of a particle is $$a=f(t)$$, then:

Kinematics

A motorcyclist who is moving along an$$x$$ axis directed toward the east has an acceleration given by $$a = (6.1 - 1.2t) m/s^{2}$$ for $$0\leq t \leq$$6.0 s. At $$t = 0$$ the velocity and position of the cyclist are 2.7 m/s and 7.3 m. (a) What is the maximum speed achieved by the cyclist? (b) What total distance does the cyclist travel between $$t = 0$$ and 6.0 s?

Kinematics

If the velocity of a particle is given by $$y={(180-16x)}^{\frac{1}{2}}$$m/s, then its acceleration will be

Kinematics

A particle is moving along the locus $$y=k-x(k>0)$$ with a constant speed $$v$$. At $$t=0$$ is at the origin and about to enter the first quadrant of $$X-y$$ axes. At some later time $$t>0, v_x = v_y$$. At this moment $$[a_y -a_x]$$.

Kinematics

A particle is moving along the $$x$$-axis whose instantaneous speed is givn by $$v^2=108-9x^2$$. The acceleration of the particle is

Kinematics

The deceleration experienced by a moving motor boat, after its engine is cut-off is given by $$\dfrac{dv}{dt}=-kv^3$$, where $$k$$ is constant. If $$v_0$$ is the magnitude of the velocity at cut-off, the magnitude of the velocity at a time t after the cut-off is

Kinematics

A point moves in a straight line so that its displacement $$x$$ metre at time t second is given by $$x^2=1+t^2$$. Its acceleration in $$ms^{-2}$$ at time t second is

Contact Details