Question

What will be the anser?

Answer 1

`5.7 m/s`

Explanation:

Here,considering speed to be the magnitude of linear velocity.

So,if it is moving with constant speed of `v` at a certain point of time,then its Centripetal acceleration is `|vec a_c|=v^2/r` where `r` is the radius of the circular path.

If its linear acceleration is `vec a_r` then,

`vec a_c + vec a_r =vec a`

Given, `|vec a|=15 ms^-2`

Again angle between `vec a_c` and `vec a_r` is `90^@`

So, `a^2=a_r^2 + a_c^2`

Or, `a^2=a_r^2 + v^4/r^2... 1`

Given, `a=15 ms^-2,r=2.5m`

If, `vec a` makes an angle of `30^@` w.r.t `vec a_c` then,

`tan 30=|vec a_r| /|vec a_c|`

I.e `1/sqrt(3) =a_r/(v^2/r)`

Given, `r=2.5m`

So, putting the values and arranging we get,

`a_r =v^2/(2.5 sqrt(3))`

Putting this value of `a_r` in the previous equation `1` we get,

`v=5.7 m/s`

Answer 2

Slightly shorter approach, same answer

Acceleration vector in non-uniform circular motion is:

• `bba(t) = underbrace(r dot theta \ bbhat e_theta)_("tangential ") - underbrace(v^2/r\ bb hat e_r)_("inward radial: " = bba_r)`

In given geometry:

`abs(bba_r) = abs(bba ) cos 30`

`implies v^2 /2.5 = 15 sqrt3/2`

`:. "speed" = abs(bbv) ~~ 5.7 " m/s"`