Kinematics & Forces

11 methods · 17 questions

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A motion graph is never read as one shape: split it at every kink into straight sections, then apply the two readings that carry all of the information.

Trigger: any velocity-time, speed-time or distance-time graph in the stem.

Instances:
(i) on a velocity-time graph the gradient of a section is the acceleration and the area under that section is the displacement over it;
(ii) compute each area as a rectangle, a triangle $\frac{1}{2} bh$ or a trapezium $\frac{1}{2} (v_{1} + v_{2}) t$ , then add them, never treat the whole graph as one block;
(iii) on a distance-time graph the gradient is the velocity, so a straight section means constant velocity and a concave-up section means acceleration;
(iv) read the two endpoints of a straight section off the grid and take the difference quotient, never one point divided by its own time.

Linked questions (3)

NSAA 2016 Section 1 Part B (ESAT Physics) · Q32

The diagram shows the velocity-time graph for an object travelling in a straight line over a period of

30 s

.

What total distance did the object travel in the

30 s

, how far from its starting position was it at the end of the

30 s

, and what was its average speed over the

30 s

A velocity-time graph. The vertical axis is velocity in m/s, marked at 2.0, 4.0, 6.0 and 8.0 above the time axis and at -2.0 below it. The horizontal axis is time in s, marked at 10, 20 and 30. A straight line starts at (0, 8.0), falls steadily to cross the time axis at (20, 0), and continues falling at the same steady rate to (30, -2.0), where a dashed line drops down to mark that final value on the vertical axis below the origin.

A $90 m$ , $70 m$ , $3.0 m s^{- 1}$
B $90 m$ , $70 m$ , $5.0 m s^{- 1}$
C $90 m$ , $90 m$ , $3.0 m s^{- 1}$
D $90 m$ , $90 m$ , $5.0 m s^{- 1}$
E $180 m$ , $140 m$ , $5.0 m s^{- 1}$
F $180 m$ , $140 m$ , $6.0 m s^{- 1}$
G $180 m$ , $180 m$ , $5.0 m s^{- 1}$
H $180 m$ , $180 m$ , $6.0 m s^{- 1}$

Worked solution

Answer:A

From

t = 0

t = 20 s

the velocity falls in a straight line from

8.0 m s^{- 1}

0

: this is a forward (positive) displacement equal to the area of that triangle,

\frac{1}{2} \times 20 \times 8.0 = 80 m

.

From

t = 20 s

t = 30 s

the velocity continues to fall in a straight line, from

0

- 2.0 m s^{- 1}

: this is a backward (negative) displacement of magnitude equal to the area of that second triangle,

\frac{1}{2} \times 10 \times 2.0 = 10 m

.

Total distance travelled (the magnitudes of both stages added together, since distance is never negative)

= 80 + 10 = 90 m

.

Displacement from the starting position at

t = 30 s

(the forward stage minus the backward stage, since the second stage carries the object back the way it came)

= 80 - 10 = 70 m

.

Average speed over the full

30 s = \frac{total distance}{total time} = \frac{90}{30} = 3.0 m s^{- 1}

.

The answer is A.

NSAA 2017 Section 1 Part B (ESAT Physics) · Q19

A car is travelling along a horizontal road in a straight line.

The graph is a velocity-time graph for part of the car's journey.

During this part of the journey, what is the total distance that the car travels while it is decelerating?

A velocity-time graph. The vertical axis is velocity in m/s, marked at 5, 10, 15, 20, 25, 30 and 35. The horizontal axis is time in s, marked every 10 s from 0 to 150. The line starts at (0, 25), stays flat at 25 until t = 30, then rises steadily to (50, 30), stays flat at 30 from t = 50 to t = 110, then falls steadily to (130, 20), then stays flat at 20 from t = 130 to t = 150.

A $400 m$
B $500 m$
C $550 m$
D $600 m$
E $750 m$
F $1400 m$
G $1800 m$
H $1900 m$

Worked solution

Answer:B

The car decelerates during the last sloped section of the graph, from

t = 110 s

t = 130 s

, where its velocity falls in a straight line from

30 m s^{- 1}

20 m s^{- 1}

.

The distance travelled during this stage is the area of the trapezium under that section of the graph:

d = \frac{1}{2} (30 + 20) \times (130 - 110) = \frac{1}{2} \times 50 \times 20 = 500 m .

The answer is B.

NSAA 2018 Section 1 Part B (ESAT Physics) · Q22

The graph is the speed-time graph for a bus travelling in a straight line between two stops.

What is the average speed of the bus during this time?

A speed-time graph on grid paper. The vertical axis is speed in m/s, gridlines every 2 from 0 to 14. The horizontal axis is time in s, gridlines every 10 from 0 to 120. A straight line rises from (0, 0) to (20, 12), stays flat at 12 from t = 20 to t = 80, then falls in a straight line from (80, 12) to (120, 0).

A $3.0 m s^{- 1}$
B $4.5 m s^{- 1}$
C $6.0 m s^{- 1}$
D $8.0 m s^{- 1}$
E $9.0 m s^{- 1}$
F $11 m s^{- 1}$
G $12 m s^{- 1}$

Worked solution

Answer:E

Total distance travelled is the area under the speed-time graph, taken in three stages:
-

0

20 s

: speed rises in a straight line from

0

12 m s^{- 1}

, area

= \frac{1}{2} \times 20 \times 12 = 120 m

.
-

20

80 s

: constant

12 m s^{- 1}

, area

= 60 \times 12 = 720 m

.
-

80

120 s

: speed falls in a straight line from

12 m s^{- 1}

0

, area

= \frac{1}{2} \times 40 \times 12 = 240 m

.

Total distance

= 120 + 720 + 240 = 1080 m

, over a total time of

120 s

.

Average speed

= \frac{1080}{120} = 9.0 m s^{- 1}

.

The answer is E.