Velocity-Time Graphs — Practice Problem Solutions

1. Displacement $=$ area under the graph $=(10\text{ m/s})(4\text{ s})=40\text{ m}$.

2. $a=\dfrac{15-0}{3}=5\text{ m/s}^2$.

3. Displacement $=$ area of the triangle $=\tfrac{1}{2}(3)(15)=22.5\text{ m}$.

4. With constant $a=+2\text{ m/s}^2$ and $v_i=0$: $v=2t$, so $v=10\text{ m/s}$ at $t=5\text{ s}$. The v-t graph is a straight line rising from the origin to the point $(5\text{ s},10\text{ m/s})$.

5. $a=\dfrac{0-20}{4}=-5\text{ m/s}^2$. The matching a-t graph is a flat horizontal line at $-5\text{ m/s}^2$ for the full 4 seconds.

6. a) Rectangle area $=(8)(3)=24\text{ m}$; triangle area $=\tfrac{1}{2}(2)(8)=8\text{ m}$; total displacement $=24+8=32\text{ m}$. b) For the first 3 s the object moves at a constant 8 m/s; for the next 2 s it decelerates steadily from 8 m/s down to a stop. c) The a-t graph is flat at $0$ for the first 3 s, then flat at $\dfrac{0-8}{2}=-4\text{ m/s}^2$ for the last 2 s.

7. A straight-line x-t graph with constant slope means the velocity itself is constant — so the matching v-t graph should be a flat, horizontal line at that value, not a line with its own positive slope. A rising v-t line would mean the velocity is increasing over time (accelerating), which contradicts the constant velocity the x-t graph actually shows.

8. (B) — the slope between $t=4.9\text{ s}$ and $t=5.1\text{ s}$ uses the narrowest window around $t=5\text{ s}$, giving the closest approximation to the true instantaneous acceleration at that exact instant.