An object moves so that $s(t) = 2t^3 - 5t^2 + 4$, in metres after $t$ seconds. Find its velocity and acceleration at $t = 3$.

$v(3) = 24\ \text{m/s}$, $a(3) = 26\ \text{m/s}^2$

A ball's height is $h(t) = -4.9t^2 + 19.6t + 2$ metres after $t$ seconds. Find its velocity at $t = 1$ and its maximum height.

$v(1) = 9.8\ \text{m/s}$; maximum height $21.6\ \text{m}$ at $t = 2\ \text{s}$

A particle moves along a line with $s(t) = t^3 - 6t^2 + 9t$, $t \ge 0$. When is it at rest, and when is it moving in the positive direction?

At rest at $t = 1$ and $t = 3$; moving in the positive direction for $0 \le t 3$.

For $s(t) = t^3 - 6t^2 + 9t$, $t \ge 0$, find when the particle is speeding up and when it is slowing down.

Speeding up on $1 3$; slowing down on $0 \le t < 1$ and $2 < t < 3$.

For $s(t) = t^3 - 6t^2 + 9t$ in metres, find the displacement and the total distance travelled in the first $4$ seconds.

Displacement $4\ \text{m}$; total distance $12\ \text{m}$

Grade 12 · MCV4U · Calculus & Vectors

MCV4U: Velocity and Acceleration

If $s(t)$ gives an object's position at time $t$ , its first derivative is velocity and its second derivative is acceleration. Motion questions are a test favourite because each part asks you to read a sign.

How It Works

$v(t) = s'(t), \qquad a(t) = v'(t) = s''(t)$

Units follow the derivative: position in metres and time in seconds give velocity in $\text{m/s}$ and acceleration in $\text{m/s}^2$ .

Direction. On a line, $v > 0$ means moving in the positive direction (right, or up) and $v < 0$ means the negative direction.

At rest. The object is at rest when $v(t) = 0$ . It turns around only if $v$ changes sign there, so check the sign on each side.

Speed is $|v|$ , the size of the velocity without its direction.

Speeding up or slowing down. Compare the signs of $v$ and $a$ . Same sign: the object is speeding up. Opposite signs: it is slowing down. A negative acceleration does not by itself mean slowing down. An object moving left with negative acceleration is speeding up.

Distance and displacement. Displacement is the final position minus the starting position. Total distance counts every metre travelled, so split the time interval at each turning point, find how far it moved on each piece, and add the absolute values.

A sign chart does most of the work. Put the zeros of $v$ and of $a$ on one number line for $t \ge 0$ , test a value in each interval, and read off direction and speeding up or slowing down from the two rows.

For a ball thrown upward, $h(t) = -4.9t^2 + v_0t + h_0$ in metres. Its acceleration is the constant $-9.8\ \text{m/s}^2$ from gravity, and its maximum height is where $v = 0$ .

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
An object moves so that $s(t) = 2t^3 - 5t^2 + 4$ , in metres after $t$ seconds. Find its velocity and acceleration at $t = 3$ .
Show solution
1. $v(t) = 6t^2 - 10t$ , so $v(3) = 54 - 30 = 24$ .
2. $a(t) = 12t - 10$ , so $a(3) = 36 - 10 = 26$ .
Answer: $v(3) = 24\ \text{m/s}$ , $a(3) = 26\ \text{m/s}^2$
Easy
A ball's height is $h(t) = -4.9t^2 + 19.6t + 2$ metres after $t$ seconds. Find its velocity at $t = 1$ and its maximum height.
Show solution
1. $v(t) = -9.8t + 19.6$ , so $v(1) = 9.8\ \text{m/s}$ , still rising.
2. Maximum height when $v = 0$ : $-9.8t + 19.6 = 0$ , so $t = 2$ .
3. $h(2) = -19.6 + 39.2 + 2 = 21.6$ .
Answer: $v(1) = 9.8\ \text{m/s}$ ; maximum height $21.6\ \text{m}$ at $t = 2\ \text{s}$
Medium
A particle moves along a line with $s(t) = t^3 - 6t^2 + 9t$ , $t \ge 0$ . When is it at rest, and when is it moving in the positive direction?
Show solution
1. $v(t) = 3t^2 - 12t + 9 = 3(t - 1)(t - 3)$ , zero at $t = 1$ and $t = 3$ .
2. Test each interval: $v(0) = 9 > 0$ , $v(2) = -3 < 0$ , $v(4) = 9 > 0$ .
3. So it moves forward, turns at $t = 1$ , moves back, and turns again at $t = 3$ .
Answer: At rest at $t = 1$ and $t = 3$ ; moving in the positive direction for $0 \le t < 1$ and $t > 3$ .
Medium
For $s(t) = t^3 - 6t^2 + 9t$ , $t \ge 0$ , find when the particle is speeding up and when it is slowing down.
Show solution
1. $v(t) = 3(t - 1)(t - 3)$ : positive on $[0, 1)$ , negative on $(1, 3)$ , positive after $3$ .
2. $a(t) = 6t - 12$ : negative on $[0, 2)$ , positive after $2$ .
3. Same signs on $(1, 2)$ , both negative, and on $t > 3$ , both positive: speeding up.
4. Opposite signs on $[0, 1)$ and $(2, 3)$ : slowing down.
Answer: Speeding up on $1 < t < 2$ and $t > 3$ ; slowing down on $0 \le t < 1$ and $2 < t < 3$ .
Test level
For $s(t) = t^3 - 6t^2 + 9t$ in metres, find the displacement and the total distance travelled in the first $4$ seconds.
Show solution
1. Displacement: $s(4) - s(0) = (64 - 96 + 36) - 0 = 4$ .
2. The particle turns at $t = 1$ and $t = 3$ , so split there: $s(0) = 0$ , $s(1) = 4$ , $s(3) = 0$ , $s(4) = 4$ .
3. Distances: $|4 - 0| + |0 - 4| + |4 - 0| = 4 + 4 + 4$ .
Answer: Displacement $4\ \text{m}$ ; total distance $12\ \text{m}$

Learn It on Martian Lab

Martian Lab teaches this in the Higher Derivatives lesson of its MCV4U Derivatives unit. Watch it, then take the Derivatives practice tests to see if it stuck.

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MCV4U: Velocity and Acceleration

How It Works

5 Worked Examples

Learn It on Martian Lab

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