Wemazi

Topic 8.2

Recover velocity or position

Builds on using a definite integral to find accumulated change. 6 minutes.

A cart that slows to a stop

A cart rolls along a straight track with velocity \(v(t) = 6 - t\) meters per second, so it slows down and stops at \(t = 6\). At \(t = 0\), the cart is at the 4 meter mark.

The graph shows the velocity, and the number line below shows the position of the cart. Drag the orange edge to the right. The green area from 0 to \(t\) grows. The cart moves to the right by the same number of meters, because a velocity times a time is a distance.

At \(t = 4\) the area is 16, so the cart is at \(4 + 16 = 20\) meters.

Press Start at 10 meters. Every position moves 6 meters to the right, because each position is the start plus the area. The velocity graph does not change.

A rate, an area, a starting value

The numbers are gone. The graph shows a rate, which tells how fast a quantity changes. The line below shows the quantity, and the ring marks its value at the start.

Drag the orange edge. The green area is the change in the quantity, so the particle starts at the ring and moves by this change.

Move the Starting value slider. The ring and the particle move together and the area stays the same, because the rate gives only the change. The starting value fixes the position.

Press Use acceleration and velocity. The same picture now recovers a velocity from an acceleration.

Notation

  1. The position at time \(\cy{t}\) is the position at a known time \(t_0\) plus the change in position. The change is the area under the velocity graph from \(t_0\) to \(\cy{t}\). The teal arrow under the line shows the same change, from the ring to the particle.

    \[\cy{s(t)} = s(t_0) + \cb{\int_{t_0}^{t} v(u)\,du}\]

    The integration variable is \(u\), because \(t\) is already the upper limit.

  2. Acceleration is the rate of change of velocity, so the same rule recovers the velocity from the acceleration \(a(u)\).

    \[\cy{v(t)} = v(t_0) + \cb{\int_{t_0}^{t} a(u)\,du}\]

    Each integration needs one known value at one known time. This value is 0 only if the problem says so.

  3. You can also use an antiderivative \(V\) of \(v\). Every antiderivative is \(V(t) + C\), and the known value at \(t_0\) gives \(C\).

    \[\begin{gathered} s(t) = V(t) + C \\ C = s(t_0) - V(t_0) \end{gathered}\]

  4. The rate must be continuous on the interval, and it must be the derivative of the quantity.

When it fails

Suppose the problem gives no starting value. Press Try other starting values.

The rate graph and the area stay the same, but the particle can be at any of these positions. The area gives only the change, so a position needs a known value.

Example 1: velocity from acceleration

A skater has acceleration \(a(t) = 3t\) meters per second squared and velocity 5 meters per second at \(t = 0\). Find the velocity at \(t = 4\).

  1. The graph shows the acceleration, which is the rate of change of velocity, so the area under it is the change in velocity. The line below shows the velocity.

  2. Find the area under the acceleration graph from 0 to 4.

    \[\int_0^4 3t\,dt = \Bigl[\frac{3t^2}{2}\Bigr]_0^4 = \cg{24}\]

  3. The velocity at 4 seconds is the starting velocity 5 plus the change 24.

    \[v(4) = 5 + 24 = \cy{29} \text{ m/s}\]

Example 2: the known value is not at t = 0

A particle has velocity \(v(t) = 2t + 1\) meters per second. It is at the 7 meter mark at \(t = 2\). Find its position at \(t = 5\).

  1. The velocity is positive, so the particle moves to the right. The known position 7 is at the dashed line, \(t = 2\), so the area starts there and not at \(t = 0\).

  2. A tempting wrong move: Add the integral from 0 to 5 to the position 7.

    \[\begin{aligned} &7 + \int_0^5 (2t + 1)\,dt \\ &= 7 + 30 = \cr{37} \end{aligned}\]

    The integral from 0 to 5 includes the first 2 seconds, before the particle is at 7. Those seconds add the area marked 6 in red, so the red particle at 37 is 6 meters too far to the right.

  3. Integrate from 2, the time of the known value, to 5.

    \[\begin{aligned} s(5) &= 7 + \int_2^5 (2t + 1)\,dt \\ &= 7 + \Bigl[t^2 + t\Bigr]_2^5 \\ &= 7 + 24 = \cy{31} \end{aligned}\]

    As a check, \(t^2 + t + C\) with \(s(2) = 7\) gives \(C = 1\), so \(s(5) = 31\).

Practice

An elevator is 20 meters above the ground at \(t = 0\). Its velocity is \(v(t) = 12 - 3t\) meters per second. Find its height at \(t = 3\).

Drag the edge to 3 seconds. The velocity is positive, so the green area adds to the starting height.

Solution

  1. The change in height is the area under the velocity graph from 0 to 3.

    \[\begin{aligned} \int_0^3 (12 - 3t)\,dt &= \Bigl[12t - \frac{3t^2}{2}\Bigr]_0^3 \\ &= 36 - 13.5 \\ &= \cg{22.5} \end{aligned}\]

  2. Add the starting height, 20 meters.

    \[h(3) = 20 + 22.5 = \cy{42.5} \text{ meters}\]

Questions

A particle starts at the 5 meter mark. The graph shows its velocity in meters per second: it rises from 0 to 4 in 2 seconds, then falls to 0 at \(t = 6\).

  1. Where is the particle at \(t = 6\)?

    • 17 meters
    • 12 meters
    • 4 meters
    Answer

    17 meters

  2. What does the particle do between \(t = 2\) and \(t = 6\)?

    • It moves to the right, more and more slowly.
    • It moves to the left.
    • It stays at one place.
    Answer

    It moves to the right, more and more slowly.

  3. A cyclist has acceleration \(a(t) = 4t\) meters per second squared and velocity 3 meters per second at \(t = 1\). Find the velocity at \(t = 3\).

    • 19 meters per second
    • 16 meters per second
    • 21 meters per second
    Answer

    19 meters per second

Summary

You integrate the rate from the time of the known value to the time you want, and you add the result to the known value.

This works because the integral of a rate is the change in the quantity, and a quantity equals its known value plus its change.

You use it to find a velocity from an acceleration, or a position from a velocity, when a starting value is known.

This is one of four skills in topic 8.2 of Which & Why: AP Calculus.