Question

A particle is rotates in a circular path of radius 54m with varying speed v=4t^(2) . Here v is in m//s and t inn second . Find angle between velocity and accelearation at t=3s .

Answer 1

The angle between the velocity and acceleration will be 45°.

Explanation:

To find the value of angle between velocity and acceleration ..plz use attached figure.

We have Radial acceleration using figure

$a_{r}$ = $\frac{v^{2} }{r}$ .. given v = 4t²

then ,

$a_{r}$ = $\frac{(4t^{2} )^{2} }{r}$  given r = 54m

then

$a_{r} = \frac{8t^{4} }{27}$.........(1)

For tangential acceleration

$a_{} _{t}$ = $\frac{dv}{dt} = \frac{d(4t^{2}) }{dt}$

$a_{} _{t}$ = 8t.........(2)

At t = 3sec

$a_{r} = \frac{8x(3^{4}) }{27}$  using equation (1)

$a_{r}$ = 24m/sec²

and

$a_{t}$ = 8x3 = 24m/s² using equation (2)

For the value of angle we have

tanθ = $\frac{a_{r} }{a_{t} }$ , putting the vale then we get

tanθ = $\frac{24}{24}$ = 1

tanθ = 1

θ = tan⁻¹(1)

θ = 45°

The angle will be 45°