v(t) = t sin t^2
a(t) = t( cos t^2) (2t) + sin t^2
= 2t^2 cos t^2 + sin t^2
y(t) = (1/2) (-cos t^2) + c
given : when t = 0 , y(0) = 3
3 = (-1/2)cos 0 + c
3 = -1/2 + c
c = 3.5
so y(t) = (-1/2) cos t^2 + 3.5
You now have enough to answer all the other questions.
Remember that if the first derivative is posititive, the function is increasing, (if a(t) > 0, then v(t) increases)
y(2) = (-1/2) cos (4) + 3.5 = appr 4.8
y(0) = 3
so the distance traveled from t-0 to t-2 = 4.8-3 = 1.8
5. A particle moves along the y – axis with velocity given by v(t)=tsine(t^2) for t>=0 .
a. In which direction (up or down) is the particle moving at time t = 1.5? Why?
b. Find the acceleration of the particle at time t= 1.5. Is the velocity of the particle increasing at t = 1.5? Why or why not?
c. Given that y (t) is the position of the particle at time t and that y (0) = 3, find y (2).
d. Find the total distance traveled by the particle from t = 0 to t = 2.
1 answer