The Kinematic Equations — Practice Problem Solutions

1. $v_f=v_i+at=0+(5)(3)=15\text{ m/s}$.

2. $\Delta x=v_it+\tfrac{1}{2}at^2=(2)(3)+\tfrac{1}{2}(4)(3)^2=6+18=24\text{ m}$.

3. (a) $a=\dfrac{30-10}{4}=5\text{ m/s}^2$. (b) $\Delta x=\dfrac{v_i+v_f}{2}t=\dfrac{10+30}{2}(4)=80\text{ m}$.

4. Using $v_f^2=v_i^2+2a\Delta x$: $v_f^2=0+2(3)(54)=324\implies v_f=18\text{ m/s}$. Then $t=\dfrac{v_f-v_i}{a}=\dfrac{18}{3}=6\text{ s}$.

5. (a) $t=\dfrac{v_f-v_i}{a}=\dfrac{0-15}{-3}=5\text{ s}$. (b) $\Delta x=\dfrac{v_i+v_f}{2}t=\dfrac{15+0}{2}(5)=37.5\text{ m}$.

6. a) $v_f=at=(1.5)(20)=30\text{ m/s}$. b) $\Delta x=\tfrac{1}{2}at^2=\tfrac{1}{2}(1.5)(20)^2=300\text{ m}$. c) Since $\Delta x=\tfrac{1}{2}at^2$, distance depends on time squared. Doubling the time doesn't double the distance — it quadruples it: $\tfrac{1}{2}(1.5)(40)^2=1200\text{ m}$, which is four times 300 m, not two times. d) Doubling the acceleration instead (same 20 s) gives $\tfrac{1}{2}(3.0)(20)^2=600\text{ m}$ — exactly double the original 300 m. This IS different from part (c): distance scales with the square of time, but only scales linearly with acceleration.

7. True. Both objects start from rest at the same position and instant, so $\Delta x=\tfrac{1}{2}at^2$ applies to each with the same $t$. Since $t^2$ is identical for both and $a$ appears as a simple multiplier, whichever object has the larger $a$ produces a strictly larger $\Delta x$ for any given $t>0$ — there's no way for the smaller acceleration to "catch up" in distance.