SME
Grade 8 Science · Problem type
A walk on a position-time graph
Use this for “On a position-time graph, when is the walker standing still, moving fastest, or heading back to the start?”
Chart
Slope = rise ÷ run = 3 m ÷ 2 s = 1.5 m/s
d = v × t = 1.5 m/s × 2 s = 3 m
A flat part stands still
A part that falls goes back toward the start
This is an example. Type your own values to start a new problem.
t
Time of the first leg
example
v
Speed on the first leg
example
d
Distance after the first leg
calculated
Formulas
d
=
v
×
t
(distance
after
the
first
leg
=
speed
on
the
first
leg
×
time
of
the
first
leg)
3 = 1.5 × 2
Assumptions
•
A walker starts at the 0 m mark and records her position every second.
•
Her first leg is yours to set. Then she walks at 0.5 m/s for 4 s, stands still for 3 s and walks back at 0.4 m/s for 5 s.
•
Steepest part: fastest. Flat part: standing still. Falling part: walking back toward 0.
Step-by-step
WE KNOW
v = 1.5 m/s (speed on the first leg), t = 2 s (time of the first leg)
FIND
distance after the first leg (d)
Step 1 · Find distance after the first leg (d)
Use d = v × t (distance after the first leg = speed on the first leg × time of the first leg)
Each
second
covers
v
meters:
multiply
by
the
seconds.
d
=
1.5
×
2
d
=
3
m
Check
Put the numbers back in. Both sides should match.
3 = 1.5 × 2 ✓
Topic check
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