"A particle moves along the x-axis so that its velocity at time T is given by:
v(T)= -(T+1)sin(T^2/2)
At T=0 the particle is at position x=1.
a) Find the acceleration of the particle at time T=2. Is the speed of the particle increasing at T=2? Why or why not?
b) Find all times T in the open interval 0<T<3 when the particle changes direction. Justify your answer. "
So I know to get the acceleration you have to derive the velocity function. This derivation consists of product quotient and chain rule.
So I got
V1(x)= Sin(t^2/2)*(-1)-(t+1)(cos(t^2/2))*(2*2t/4)
Then you would just substitute in 2 for t to find the acceleration at t=2 right? Is this the correct derivative of the velocity function, and how would you go about part b?
Thank you!
Ok so for part A I did:
a(2)= -(2+1)cos(2^2/2)*2-sin(2^2/2)
a(2)= -4.083.
The speed of the particle is decreasing because the acceleration is negative. Correct?
And for part B I did:
V(t) -(t+1)sin(t^2/2)=0
V(0)= -(0+1)sin(0^2/2)=0
V(1)= -(1+1)sin(1^2/2)= -1.438
V(2) = -(2+2)sin(2^2/2)= -3.637
V(3)= -(3+2)sin(3^2/2) = 4.887
The particle changes direction when t=3 because the sign goes from negative to positive.
Are these answers correct?