SME
Calculus I · Problem type
Position from a growing acceleration
Use this for “A cart’s acceleration is a(t) = 6t m/s². It starts at s = 1 m with v = 2 m/s. Find v and s at t = 2 s.”
Chart
s(2) = 6
Tangent at t = 2: slope s′(2) = 5, y = 5t − 4
This is an example. Type your own values to start a new problem.
a(t)
=
k (m/s³)
×
t
v(0)
=
v₀ (m/s)
,
s(0)
=
s₀ (m)
v(
t (s)
)
=
v (m/s)
,
s(
t (s)
)
=
s (m)
Formulas
v
=
v₀
+
k
×
t
2
2
5
=
−1
+
3
×
2
2
2
s
=
s₀
+
v₀
×
t
+
k
×
t
3
6
6
=
4
+
(−1)
×
2
+
3
×
2
3
6
Assumptions
•
The acceleration grows steadily from 0: a(t) = kt, where the jerk k is how much it grows each second.
•
Integrate the acceleration to get v and the velocity to get s. Each integration adds a constant, fixed by the start values v(0) = v₀ and s(0) = s₀.
•
Position, velocity and acceleration are signed along one line: + one way and − the other.
Step-by-step
WE KNOW
k = 3 m/s³, v₀ = −1 m/s, s₀ = 4 m, t = 2 s
FIND
velocity at time t (v), position at time t (s)
Step 1 · Find velocity at time t (v)
Use v = v₀ + k × t² ÷ 2
Integrate
the
acceleration
kt:
v
(
t
)
=
kt
²
2
+
C.
At
t
=
0
that
leaves
C,
so
the
constant
is
v₀
.
v
=
−1
+
3
×
2
2
2
v
(
t
)
=
1.5
t
²
−
1
v
=
5
m/
s
Step 2 · Find position at time t (s)
Use s = s₀ + v₀ × t + k × t³ ÷ 6
Integrate
the
velocity
v₀
+
kt
²
2
:
s
(
t
)
=
v₀
t
+
kt
³
6
+
C.
At
t
=
0
that
leaves
C,
so
the
constant
is
s₀
.
s
=
4
+
(−1)
×
2
+
3
×
2
3
6
s
(
t
)
=
0.5
t
³
−
t
+
4
s
=
6
m
Check
Put the numbers back in. Both sides should match.
5
=
−1
+
3
×
2
2
2
✓
6
=
4
+
(−1)
×
2
+
3
×
2
3
6
✓
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An initial-value problem with f′ = A·eᵏˣ
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