AP Physics C: Mechanics

Unit 1

Kinematics

8 practice questions · Free response

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Notes

Calculus definitions

$$v(t) = \frac{dx}{dt} \qquad a(t) = \frac{dv}{dt} = \frac{d^2x}{dt^2}$$
$$x(t) = x_0 + \int_0^t v\,dt \qquad v(t) = v_0 + \int_0^t a\,dt$$
  • The constant-acceleration equations only apply when $a$ is constant. If $a$ depends on time, integrate.
  • If $a$ depends on velocity, separate variables: $\displaystyle \frac{dv}{dt} = f(v) \Rightarrow \int\frac{dv}{f(v)} = \int dt$.
  • If $a$ depends on position, use $\displaystyle a = v\frac{dv}{dx}$.

Vectors and 2D motion

  • Position $\vec r = x\hat i + y\hat j$; velocity and acceleration are derivatives of each component.
  • Projectile motion: $a_x = 0$, $a_y = -g$. Range on level ground: $\displaystyle R = \frac{v_0^2\sin 2\theta}{g}$, largest at 45°.
  • Uniform circular motion: $\displaystyle a_c = \frac{v^2}{r}$ toward the center. With changing speed, there's also tangential acceleration $\displaystyle a_t = \frac{dv}{dt}$.

Relative motion

$\vec v_{A/C} = \vec v_{A/B} + \vec v_{B/C}$.

Practice

A particle's position is $x(t) = 4t^3 - 2t$ (m). What is its velocity at $t = 1$ s?
For the same particle, what is its acceleration at $t = 1$ s?
An object starts from rest with $a(t) = 6t$ (m/s²). What is its velocity at $t = 2$ s?
For the same object starting at $x = 0$, what is its position at $t = 2$ s?
At what launch angle is the range of a projectile on level ground greatest (no air resistance)?
A car moving in a circle speeds up. Its acceleration vector points:
The velocity of an object is $v(t) = 3t^2 - 12$ (m/s). When is it momentarily at rest for $t > 0$?
A projectile is launched at 20 m/s and 30°. With $g = 10$ m/s², what is its range on level ground?

Free response

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Free-response question

A particle moves along the $x$-axis with velocity $v(t) = t^2 - 6t + 8$ (m/s) for $t \ge 0$, starting at $x = 0$.

(a) Find when the particle is at rest.

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$t^2 - 6t + 8 = (t - 2)(t - 4) = 0$, so $t = 2$ s and $t = 4$ s.

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(b) Find the particle's position at $t = 3$ s.

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$\displaystyle x(3) = \int_0^3 (t^2 - 6t + 8)\,dt = \left[\frac{t^3}{3} - 3t^2 + 8t\right]_0^3 = 9 - 27 + 24 = 6$ m.

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(c) Find the acceleration at $t = 3$ s, and say whether the particle is speeding up or slowing down then.

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$a = 2t - 6 = 0$ at $t = 3$. At that instant the acceleration is zero, so the speed is momentarily neither increasing nor decreasing (it's at its maximum speed of 1 m/s while moving in the negative direction).

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