Calculus question

Question

For t≥0, a particle moved along the x-axis. The velocity of the particle at time t is given by …v(t)=1+2cos((t²)/2)… the particle is at position x=2 at time t=4.

a. At time t=4, is the particle speeding up or slowing down? Justify your answer.

b. Find all times t in the interval 0<t<3 when the particle changes direction. Justify your answer.

c. Find the position of the particle at time t=0.

Answer 1

Solution:

The velocity v(t) = 1 + 2cos(t^2/2)

The acceleration a(t) = v'(t) = -2tsin(t^2/2)

(a)

At t = 4, v(4) = 1 + 2cos(4^2/2) ≈ 0.709 and a(4) = -2tsin(4^2/2) ≈ -7.915.

Since v(4) and a(4) have different signs, the particle is slowing down.

(b)

Let v(t) = 1 + 2cos(t^2/2) = 0. Then

cos(t^2/2) = -1/2

When 0 < t < 3, 0 < t^2/2 < 4.5.

So t^2/2 = 2pi/3 or 4pi/3

t ≈ 2.047 or 2.894

Now v(1) > 0, v(2.5) <0, v(3) > 0,

So v(t) changes the sign at t = 2.047 and 2.894.

Therefore, the particle changes the direction at t = 2.047 and t = 2.894.

(c)

We know v(t) = x'(t) = 1+ 2cos(t^2/2).

Then x(4) - x(0) = int_0^4 [1+ 2cos(t^2/2)] dt ≈ 0.629 (by FTC)

So x(0) ≈ x(4) - 0.629

= 2 - 0.629

= 1.731

The position of the particle at t = 0 is x = 1.731.