Displacement & Distance Traveled of v(t) = t^3 - 5t^2 on the interval [-1,6]

Question

The velocity function (in meters per second) for a particle moving along a line is given by v(t) = t³-5t². Find the displacement and the distance traveled by the particle during the time interval [-1,6].

I have attempted this problem multiple times, but keep getting the incorrect answer. I am unsure where I am going wrong.

Answer 1

Displacement is just the integral of v. But distance is, by definition, the integral of |v|.

The trouble is that |v| is piecewise, i.e. |v| = -v when v<0. So solve v<0 and you get t<5. Now just calculate.

Here is the most direct and complete way to do that (I'm including extra steps for your sake):

∫_-1⁶|v|dt

= ∫_-1⁵ |v| dt + ∫₅⁶ |v| dt

= ∫_-1⁵ v dt + ∫₅⁶ (-v) dt

= ∫_-1⁵ v dt - ∫₅⁶ v dt

= [V(5) - V(-1)] - [V(6) - V(5)]

= 2V(5) - V(-1) - V(6)

Nothing is left but finding the antiderivative:

V(t) = t⁴/4 - 5t³/3

Now you can finish the last line of the calculation above.

Answer 2

s(t) = position at time t

v(t) = velocity at time t = s'(t)

Displacement = change in position = s(6) - s(-1) = ∫_{(-1 to 6)} (t³ - 5t²)dt = [(1/4)t⁴ - (5/3)t³]_{(-1 to 6)} = -36

From t = -1 to t = 6, the particle moves 36 meters to the left.

v(t) = t³ - 5t² = t²(t - 5) = 0 when t = 0 or t = 5.

When -1 < t < 0, v(t) < 0. So, the particle is moving to the left.

When 0 < t < 5, v(t) < 0. So, the particle is moving to the left.

When 5 < t < 6, v(t) > 0, So, the particle is moving to the right.

Total distance = ∫_{(-1 to 6)} l v(t) l dt = [(5/3)t³ - (1/4)t⁴]_{(-1 to 5)} + [(1/4)t⁴ - (5/3)t³]_{(5 to 6)}

Evaluate to find the total distance.

Answer 3

Displacement is the overall change in location, so it's the integral from -1 to 6 of the function.

Distance is total change of location, regardless of direction. You can either graph or set the function = 0. If you graph it, you'll see it's negative from -1 to 5, and it's positive from 5 to 6.

Take the absolute value of the integrals of the function from -1 to 5 and from 5 to 6. Add them. That's the distance.