Question

The acceleration of a particle is given by a(t)=cos(t)-sin(t). At t=0, the velocity is 1. During the time from pi < t< 2pi, what is the acceleration of the particle when the velocity is 0.


The less than signs are supposed to be less than or equal to signs.


Please help. Thank you.

Answer 1

Use <= for less than or equal.

a(t) = cos(t)-sin(t)
v(t) = sin(t)+cos(t) + c
v(0)=1, so c=0
v=0 when sin(t)+cos(t)=0.
That is, sin(t) = -cos(t)
sin(t)=cos(t) when t=π/4, so using that as a reference angle, that makes t = 3π/4 or 11π/4

Only 11π/4 is between π and 2π.
So, a(11π/4) = 1/√2 - (-1/√2) = √2