Question

At any instant of time acceleration and velocity of a particle are given by v=6i+8j and a=i+j. Find rate of change of speed at that instant.

Answer 1

The rate of change of speed at that instant is 1.4 m/s

Explanation:

If the motion is in straight line

Then the rate of change of speed is the magnitude of acceleration

Here we cannot say if the motion is rectilinear

Given

At any instant

The velocity of particle

$\vec v=6\hat i+8\hat j$

Here, x-component of velocity $v_x=6$

And y-component of velocity $v_y=8$

Acceleration of the particle

$\vec a=\hat i+\hat j$

Here, x-component of acceleration $a_x=1$

And y-component of acceleration $a_y=1$

speed at any instant

$v=\sqrt{v_x^2+v_y^2}$

$\implies v^2=v_x^2+v_y^2$

$\implies 2v\frac{dv}{dt}=2v_x\frac{dv_x}{dt}+2v_y\frac{dv_y}{dt}$   (Differentiating the equation w.r.t. t)

$\implies v\frac{dv}{dt}=v_xa_x+v_ya_y$   (dv/dt is rate of change of speed)

$\implies \sqrt{6^2+8^2}\frac{dv}{dt}=6\times 1+8\times 1$

$\implies \sqrt{100}\frac{dv}{dt}=14$

$\implies \frac{dv}{dt}=\frac{14}{10}$

$\implies \frac{dv}{dt}=1.4$ m/s

Hope this answer is helpful.

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Answer 2

Explanation:

as there is change in magnitude of velocity(speed) tangential acceleration will occur