Acceleration — Practice Problem Solutions

1. $a=\dfrac{30-10}{5}=4\text{ m/s}^2$.

2. $a=\dfrac{2-8}{3}=-2\text{ m/s}^2$ (negative, since velocity is decreasing).

3. $a=\dfrac{-12-12}{2.4}=\dfrac{-24}{2.4}=-10\text{ m/s}^2$.

4. $v_f=v_i+at=0+(6)(4)=24\text{ m/s}$.

5. $t=\dfrac{v_f-v_i}{a}=\dfrac{0-15}{-3}=5\text{ s}$.

6. a) Interval A: $a=\dfrac{25-0}{5}=5\text{ m/s}^2$. Interval B: $a=0$ (constant velocity). Interval C: $a=\dfrac{0-25}{2.5}=-10\text{ m/s}^2$. b) Interval C has the largest-magnitude acceleration ($10\text{ m/s}^2$, versus $5\text{ m/s}^2$ for A and $0$ for B). c) The classmate is wrong — zero acceleration just means velocity isn't changing, not that the car isn't moving. During Interval B the car is still cruising along at a steady 25 m/s; "nothing happening" describes rest, not constant-velocity motion.

7. The acceleration stays negative (downward) the entire time because it's caused by gravity, which pulls downward no matter which way the object happens to be moving at that instant. Acceleration describes the direction of the net force, not the direction of the current velocity — so it doesn't flip sign just because the object turns around and starts falling instead of rising.