A particle starts from rest at t = 0 with acceleration a(t) = 6t m/s² (where t is in seconds). Find: (a) the velocity at t = 3 s, (b) the position at t = 3 s, and (c) the average velocity over the interval [0, 3] s.
Solving kinematics problems when acceleration depends on time, position, or velocity
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Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 3 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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Motion with Variable Acceleration is part of the AP Physics C: Mechanics course on Study Mondo, specifically in the Kinematics section. You can explore the full course for more related topics and practice resources.
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∫
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t′
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d
t′
x(t)=x0+∫t0tv(t′)dt′
Example: Linear Time Dependence
If a(t)=At+B where A and B are constants:
v(t)=v0+∫0t(At′+B)dt′=v0+21At2+Bt
x(t)=x0+∫0t(v0+21Atx0+v0t+61At3+21Bt2
Acceleration as a Function of Position: a(x)
When acceleration depends on position, use the chain rule:
a=dtdv=dxdvdtdx=vdxdv
Therefore:
vdv=a(x)dx
Integrating both sides:
∫v0vv′dv′=∫x0xa(x′)dx′
21(v2−v02)=∫x0xa(x′)dx′
Example: Spring Force
For a spring with a=−mkx:
vdv=−mkxdx
∫v0vv′dv′=−mk∫x0xx
21(v2−v02)=−2mk(x2−x02)
v2=v02−mk(x2−x02)
Acceleration as a Function of Velocity: a(v)
When acceleration depends on velocity:
a=dtdv=a(v)
Separate variables:
a(v)dv=dt
Integrate:
∫v0va(v′)dv′=∫0tdt′=t
Example: Linear Drag Force
For drag force a=−bv (where b>0):
dtdv=−bv
vdv=−bdt
∫v0vv′dv′=−b∫0tdt′
lnv0v=−bt
v(t)=v0e−bt
To find position, integrate velocity:
x(t)=∫0tv0e−bt′dt′=bv0(1−e−bt)
Quadratic Drag Force
For drag proportional to v2: a=−bv2
dtdv=−bv2
∫v0vv′2dv′=−b∫0tdt′
−v1+v01=−bt
v(t)=1+bv0tv0
Falling with Air Resistance
Terminal velocity occurs when drag force equals gravitational force:
For linear drag: a=g−bv
At terminal velocity: vt=bg
General solution:
v(t)=vt(1−e−bt)+v0e−bt
For quadratic drag: a=g−bv2
Terminal velocity: vt=bg
v(t)=v0+∫0ta(t′)dt′=∫0t6t′dt′
v(t)=6⋅2t2=3t2
v(3)=3(3)2=27 m/s
(b) Position at t = 3 s:
x(t)=x0+∫0tv(t′)dt′=∫0t3t′2dt′
x(t)=3⋅3t3=t3
x(3)=(3)3=27 m
(c) Average velocity:
vavg=3−0x(3)−x(0)=327
vavg=9 m/s
2Problem 2hard
❓ Question:
A rocket experiences acceleration a = (40 - 5t) m/s² until its fuel runs out. The rocket starts from rest at t = 0. Find: (a) when the fuel runs out (when a = 0), (b) the maximum velocity, and (c) the total distance traveled while fuel is burning.
💡 Show Solution
Given:
a(t) = 40 - 5t m/s²
v₀ = 0, x₀ = 0
(a) When fuel runs out:
Set a = 0:
40−5t=0
t=8 s
(b) Maximum velocity:
v(t)=∫0t(40−5t
At t = 8 s:
v(8)=40(8)−25(64)=
vmax=160 m/s
(c) Total distance:
x(t)=∫0t(40t
x(t)=40⋅2t
At t = 8 s:
x(8)=20(64)−65(512)=
x=853 m
3Problem 3hard
❓ Question:
A particle moves with position-dependent acceleration a = -kx, where k = 4 s⁻². If v = 8 m/s when x = 0, find: (a) the velocity as a function of position, (b) the maximum displacement, and (c) identify the type of motion.
💡 Show Solution
Given:
a = -kx where k = 4 s⁻²
At x = 0: v = 8 m/s
(a) Velocity as function of position:
Using a=vdxdv:
vdxdv=−kx
vdv=−kxdx
Integrating:
2v2=−2
At x = 0, v = 8:
C=264=32
2v2=−2
v2=64−4x2
v=64−4x2
(b) Maximum displacement:
At maximum displacement, v = 0:
0=64−4xmax2
xmax2=16
xmax=4 m
(c) Type of motion:
a=−kx=−(4)x
This is Simple Harmonic Motion (SHM)!
ω2=k=4⟹ω=2 rad/s
The particle oscillates with amplitude A = 4 m and angular frequency ω = 2 rad/s.
General solution: x(t)=Asin(ωt+ϕ)
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Yes, this page includes 3 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.