Distance-Time GrAphs
Summary
• The gradient of the slope gives the velocity
rise ∆d
=
, this is
• Average speed = gradient =
run ∆t
d
shown by the formula v =
t
• The unit for distance is meters (m) and time is
seconds (s) and velocity is meters per second
m
( ms −1 or )
s
For a complete tutorial on this topic visit www.learncoach.co.nz
Old NCea Questions
1. A parachutist jumps from a plane. The parachutist
falls through a distance of 2400 m during the first
60 seconds. Calculate the average speed of the
parachutist during this time.
2. A child plays with a remote control car on concrete.
a. The car starts from rest and travels a distance of
b.
6 m in 3 seconds. Calculate the average speed of
the car in the 3 seconds.
The car then travels a further 28 m at a constant
speed of 4 m s–1 for 7 seconds.
Using the information given in (a) and above,
draw in the appropriate shaped lines on the
distance-time graph to the right to represent the
journey of the car during the first 10 seconds.
Practice Questions
3. John was skating to the dairy. The graph below shows
his journey.
a. What distance had John travelled at seven
seconds?
b. In which part of his journey, A, B, or C, is John
going the fastest?
c. Give ONE reason for your answer to (b) above.
d. Calculate John’s average speed over the whole
10 second journey.
Page 2
Physics
4. Stephanie is watching a kitten playing. The kitten
starts from rest and travels a distance of 6 m in 3
seconds. The kitten then travels a further 28 m at a
constant speed of 4 ms-1 for 7 seconds. Using the
given information, draw in the appropriate shaped
lines on the distance-time graph below to represent
the journey of the kitten during the first 10 seconds.
8. Leah was paying with a toy car. She rolled it down a
slope and it took ten seconds to reach the bottom.
What was the average speed of the car? Show all
working.
9. Philip enters a downhill snowboard race. The
distance-time graph below shows his progress.
5. Jenny offers to help her neighbour move in. At one
point Jenny is pushing a couch across a room. The
room is 18 meters across and it takes her 6 seconds
to push the couch across the room. Calculate the
average speed of the couch. Show all working. Units
are required.
6. Ben climbs a ladder 20 m high. He then slides down
a pole in 8 s. Calculate his average speed. Show all
workings in your calculation.
7. Philip was watching a toddler chase a ball. The
distance-time graph is shown below.
a. How far, in meters, does Philip travel in the first
40 seconds?
b. Calculate Philip’s average speed over the entire
race.
c. Describe the motion of Philip and his snowboard
during section C.
d. Using the graph above, show that Philip’s speed
during section B is 15.7 ms-1.
10. Kyle is playing with a remote control helicopter. It
starts sitting on the floor then hits the roof, 2.5 m
up, after 2 seconds. Calculate the average speed of
the helicopter.
a. Describe the motion of the toddler for the first
15 seconds.
b. What was the toddler’s speed during the time
15 s to 25 s?
c. Calculate the average speed of the toddler for
the time 25 s to 40 s.
Page 3
ANswers
NCEA
4.
∆d 2400
=
∆t
60
-1
Average Speed = 40 ms
2.
1. v =
a. v =
(Achieved)
d 6
= = 2 ms −1 t 3
(Achieved)
b.
NB: The curved section of the graph represents
the kitten’s acceleration. If the kitten travelled at a
constant speed for the first three seconds this would
be a straight line.
(Excellence)
(Both lines correct shape - Achieved)
Practice
3.
a. 30 m
b. Part B
c. Because the slope of the graph (which shows
the amount of distance travelled over a certain
amount of time) is steepest during B, with a
gradient of 10ms-1 OR During part A he travels
20m in 6 seconds; during part B he travels 20m
in 2 seconds; and during part C he is stationary.
d
d. v =
t
Total distance = 40 m
Total time = 10 s
average speed =
40
= 4 ms −1
10
(Achieved - two of a, b, c or d)
(Merit - three of a, b, c or d)
(Excellence - all of a, b, c and d)
5. v=
d 18
=
= 3 ms-1 t
6
(Achieved)
6. v=
d 20
=
= 2.5 ms-1 t
8
(Achieved)
7. a. The toddler is travelling at constant speed.
b. The toddler’s speed was zero.
∆d 20
=
= 1.33 ms-1 (3 sf)
c. v =
∆t 15
8. v =
d 15
=
= 1.5 ms −1 t 10
(Achieved)
9. a. 0.8 km = 800 m
d 3300
= 15 ms −1
b. v = =
t
220
c. The (parabolic) shape indicates constant
deceleration.
d 3000 − 800 2200
=
=
= 15.71 ms −1
t
180 − 40
140
d. v =
10. v =
Page 4
(Merit - two of a, b or c)
(Achieved - two of a, b, c or d)
(Merit - three of a, b, c or d)
d 2.5
=
= 1.25 ms −1 t
2
(Achieved)