Question #42a96

I'm not sure if I completely understand the question on this and I'm a bit confused by the part of the question "....and that speed increases in `10ms^(-2)`......"

If a particle is travelling in circular motion, then its centripetal acceleration is equal to `v^2/r` which in this case would be `10^2/10=10ms^-2`.

I assume this acceleration is what is being referred to in the question as "....and that speed increases in `10ms^(-2)`......"

On this basis, for any particle travelling in circular motion, the instantaneous velocity at ay point is always at 90 degrees to the acceleration vector. The acceleration vector is always directed towards the centre of the circle of travel and the velocity is always tangential.

Given
Speed of the particle moving in a circular path is `v=10ms^-1`
Tangential acceleration acting on the particle `a_t=10ms^-2`

Radius of the circular path `r=10m`
The centripetal acceleration acting on the particle radially is `a_r=v^2/r=10^2/10=10ms^-2`

The resultant acceleration acting on the particle `a=sqrt(a_r^2+a_t^2)=sqrt(10^2+10^2)ms^-2`
`=>a=10sqrt2 ms^-2`
If the resultant acceleration ‘a’ makes an angle `theta` with the tangential acceleration `a_t` then

`tantheta=a_r/a_t=10/10=1`
`:.theta=tan^(_1)1=45^@`

The direction of instanteneous veylocity vector is same as the direction of tangential acceleration.

Hence the resultant acceleration will also make `45^@` angle with the instanteneous velocity vector.