Solve motion problems with the kinematic equations
Grade 11 Science Worksheet - Challenge Me
Circle the best answer, or write your answer on the line.
A cart rolling at 12 m/s brakes steadily at 3 m/s² until it stops. Using v² = v₀² + 2ax, how far does it travel while stopping, Ari?
A cart rolling at 20 m/s brakes steadily at 5 m/s² until it stops. Using v² = v₀² + 2ax, how far does it travel while stopping, Bo?
A cart rolling at 9 m/s brakes steadily at 3 m/s² until it stops. Using v² = v₀² + 2ax, how far does it travel while stopping, Esme?
In the velocity-time graph, what is the cart's acceleration, Kira?
- 2.5 m/s²
- 2 m/s²
- 10 m/s²
- 8 m/s²
Using the velocity-time graph's start and end velocities, which working finds the distance without using the time, Luz?
Use the picture in question 4.
- 10² ÷ 2 = 50 m
- (10² − 2²) ÷ (2 × 2) = 24 m
- (10 − 2) ÷ (2 × 2) = 2 m
- (10² + 2²) ÷ 2 = 52 m
Milo says x = v₀t + ½at² works for a cart whose acceleration keeps changing. Agree or disagree?
- Agree: any equation works if t is unknown
- Agree: braking a is positive anyway
- Disagree: it needs constant acceleration
- Agree: and the ½ can be dropped
Pia says v² = v₀² + 2ax finds a stopping distance when the time isn't known. Agree or disagree?
- Disagree: it should be v₀t + at²
- Agree: it has no t in it
- Disagree: use x = v₀t + ½at², though t is unknown
- Disagree: stopping makes a positive
Rearrange v = v₀ + at to find t. Which is right, Ren?
- t = (v − v₀) ÷ a
- t = (v + v₀) ÷ a
- t = a ÷ (v − v₀)
- t = v − v₀ − a
Rearrange v² = v₀² + 2ax to find a. Which is right, Tala?
- a = 2x ÷ (v² − v₀²)
- a = (v − v₀) ÷ 2x
- a = (v² − v₀²) ÷ 2x
- a = (v² + v₀²) ÷ 2x
Up is positive. A spring toy launches a ball straight up at 14.7 m/s, with g = 9.8 m/s². How high does it rise, Zee?
- About 22 m
- About 11 m
- About 1.5 m
- About 0.75 m