SME
Grade 11 Science · Problem type
Position–time graph: velocity is the slope
Use this for “x = 10 − 6t + t². How fast is it moving at t = 5 s, where is it, and when does it turn round?”
Chart
v = v₀ + at = (−6) + 2 × 8
v = 10 m/s
x(t₁)
=
x₀
+
v₀t₁
+
1
2
at₁²
x(t₁)
=
10
+
(−6)
×
5
+
1
2
×
2
×
5
2
=
5
m
The tangent’s slope is the velocity at t₁.
v₁ = v₀ + at₁ = (−6) + 2 × 5
v₁ = 4 m/s
The tangent slopes up: moving forward (+).
Position at t₁: 5 m
This is an example. Type your own values to start a new problem.
x₀
Starting position
example
v₀
Initial velocity
example
a
Acceleration
example · m/s²
t
Time shown on the graph
example
t₁
Time of the tangent
example
v₁
Velocity at t₁ (the slope)
calculated
x₁
Position at t₁
calculated
Formulas
v₁
=
v₀
+
a
×
t₁
4 = −6 + 2 × 5
x₁
=
x₀
+
v₀
×
t₁
+
1
2
×
a
×
t₁
2
5
=
10
+
(−6)
×
5
+
1
2
×
2
×
5
2
Assumptions
•
With a steady acceleration the position is x = x₀ + v₀t + ½at², a parabola against time.
•
The slope of the tangent at a moment is the velocity at that moment.
•
Where the tangent is flat, it stops for an instant and turns round: type v₁ = 0 to find when.
•
In x = x₀ + v₀t + ½at², the number before t² is ½a, so a is twice it.
Step-by-step
WE KNOW
x₀ = 10 m, v₀ = −6 m/s, a = 2 m/s², t = 8 s, t₁ = 5 s
FIND
velocity at t₁ (the slope) (v₁), position at t₁ (x₁)
Step 1 · Find velocity at t₁ (the slope): v₁
Use v₁ = v₀ + a × t₁
The
tangent’s
slope
is
the
velocity
then:
v₀
plus
a
for
every
second
up
to
t₁
.
v₁
=
−6
+
2
×
5
v₁
=
−6
+
10
v₁
=
4
m/s
Step 2 · Find position at t₁ (x₁)
Use x₁ = x₀ + v₀ × t₁ + ½ × a × t₁²
Start
at
x₀
,
then
add
the
displacement
up
to
t₁
.
x₁
=
10
−
6
×
5
+
1
2
×
2
×
5
2
x₁
=
10
−
6
×
5
+
1
2
×
2
×
25
x₁
=
10
−
30
+
25
x₁
=
5
m
Check
Put the numbers back in. Both sides should match.
4 = −6 + 2 × 5 ✓
5
=
10
−
6
×
5
+
1
2
×
2
×
5
2
✓
Topic check
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