Speed-Time Graphs and Acceleration

Worked example

1. 1

   $0\text{ to }4\text{ s: the line goes up}$ — The speed rises, so the object is speeding up.

2. 2

   $4\text{ to }10\text{ s: flat at }12\text{ m/s}$ — A flat line means the speed stays the same. Read across to the side scale for the value.

3. 3

   $10-4=6\text{ s}$ — How long the speed was steady: end time minus start time.

4. 4

   $10\text{ to }16\text{ s: the line goes down}$ — The speed falls, so the object is slowing down.

5. 5

   $\text{at }16\text{ s the line reaches }0$ — A speed of 0 means it has stopped.

Worked example

1. 1

   $12-0=12\text{ m/s}$ — Change in speed is the end speed minus the start speed. This is the rise of the line.

2. 2

   $4-0=4\text{ s}$ — Time taken is the end time minus the start time. This is the run of the line.

3. 3

   $\frac{12}{4}$ — Change in speed on top, time taken on the bottom.

4. 4

   $=3{\text{ m/s}}^2$ — The unit is metres per second, every second: m/s².

Worked example

1. 1

   $24-10=14\text{ m/s}$ — Change in speed: end speed minus start speed. Do not use 24 alone, because the car was already moving.

2. 2

   $\frac{14}{7}=2{\text{ m/s}}^2$ — Divide by the 7 seconds the change took.

Worked example

1. 1

   $12\to 0\text{ m/s}$ — Read the speed at the start and the end of the sloping-down line.

2. 2

   $0-12=-12\text{ m/s}$ — Change in speed is end minus start. It is negative because the speed falls.

3. 3

   $16-10=6\text{ s}$ — Time taken: end time minus start time.

4. 4

   $\frac{-12}{6}=-2{\text{ m/s}}^2$ — Change in speed divided by time. A negative gradient means slowing down.

5. 5

   $\text{deceleration}=2{\text{ m/s}}^2$ — Deceleration is the size of this number. The word already means slowing down, so drop the minus sign.

Worked example

1. 1

   $0\text{ to }5\text{ s: }10÷5=2{\text{ m/s}}^2$ — Straight line going up, so the cyclist speeds up at a constant rate. The speed gained is 10 in 5 seconds.

2. 2

   $5\text{ to }25\text{ s: constant }10\text{ m/s}$ — The line is flat, so the speed does not change. Give the speed.

3. 3

   $35-25=10\text{ s}$ — The slowing part lasts from 25 s to 35 s.

4. 4

   $25\text{ to }35\text{ s: }10÷10=1{\text{ m/s}}^2$ — Straight line going down, so the cyclist slows at a constant rate. The speed lost is 10 in 10 seconds.

Worked example

1. 1

   $\text{start speed}=5\text{ m/s}$ — The car is already moving, so it starts at 5 and not at 0.

2. 2

   $1.5\times 6=9\text{ m/s}$ — Each second adds 1.5 m/s. Over 6 seconds that is 1.5 × 6.

3. 3

   $5+9=14\text{ m/s}$ — Start speed plus the speed gained is the end speed.

Worked example

1. 1

   $20-0=20\text{ m/s}$ — It must lose all its speed to reach 0.

2. 2

   $20÷4=5\text{ s}$ — Each second it loses 4 m/s, so the time is the speed to lose divided by 4.

3. 3

   $12+5=17\text{ s}$ — Braking started at 12 s. Add the 5 s to get the time it stops.

4. 4

   $\text{line from }(12,20)\text{ to }(17,0)$ — Join the two points with a straight line.