SME
Grade 11 Science · Problem type
Turning round: displacement and distance
Use this for “A ball rolls up a ramp at 12 m/s and slows at 4 m/s². Where is it after 5 s, and how far has it rolled?”
Chart
v = v₀ + at = 12 + (−4) × 5
v = −8 m/s
Δx
=
v₀ + v
2
×
t
Δx
=
12 + (−8)
2
×
5
=
10
m
It turns round at 3 s, where the velocity is 0.
Area above the axis 18 m, below −8 m: displacement 10 m, distance travelled 26 m.
Distance traveled: 26 m
This is an example. Type your own values to start a new problem.
Δx (m)
=
v₀ (m/s)
×
t (s)
+
½
×
(
a (m/s²)
)
×
(
t (s)
)
2
v
Final velocity
calculated
D
Distance traveled
worked out (not typed)
Formulas
v
=
v₀
+
a
×
t
−8 = 12 + (−4) × 5
Δx
=
v₀
×
t
+
1
2
×
a
×
t
2
10
=
12
×
5
+
1
2
×
(−4)
×
5
2
D
=
(
v₀
2
+
v
2
)/(−2
×
a
)
26
=
(
12
2
+
(−8)
2
)/(−2
×
(−4))
Assumptions
•
+ is the way it starts, so a < 0 slows it, stops it for an instant and brings it back.
•
For the distance, v₀ and v have opposite signs: it has turned round.
•
Distance adds the way out and the way back; displacement takes the way back away.
Step-by-step
WE KNOW
v₀ = 12 m/s, a = −4 m/s², t = 5 s
FIND
final velocity (v), displacement (Δx), distance traveled (D)
Step 1 · Find final velocity (v)
Use v = v₀ + a × t
Start
at
v₀
and
add
a
for
every
second.
v
=
12
−
4
×
5
v
=
12
−
20
v
=
−8
m/s
Step 2 · Find displacement (Δx)
Use Δx = v₀ × t + ½ × a × t²
The distance at the starting velocity, plus the extra that the steady acceleration adds.
Δx
=
12
×
5
+
1
2
×
(−4)
×
5
2
Δx
=
12
×
5
+
1
2
×
(−4)
×
25
Δx = 60 − 2 × 25
Δx = 60 − 50
Δx = 10 m
Step 3 · Find distance traveled (D)
Use D = (v₀² + v²)/(−2 × a)
Out
to
the
stop
is
v₀
2
−2
a
and
back
is
v
2
−2
a
,
both
positive
since
a
<
0:
add
them.
D
=
(
12
2
+
(−8)
2
)/(−2
×
(−4))
D
=
(144
+
64)/8
D
=
208
8
D
=
26
m
Check
Put the numbers back in. Both sides should match.
−8 = 12 − 4 × 5 ✓
10
=
12
×
5
+
1
2
×
(−4)
×
5
2
✓
26
=
(
12
2
+
(−8)
2
)/(−2
×
(−4))
✓
Topic check
See what you know: 5 questions on this topic
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