Purpose: Test your understanding of 1-D and 2-D motion, vectors, free fall, projectile motion, and relative velocity. Instant feedback with worked solutions on every question.
📋 Formula Reference
\( v = v_0 + at \) | \( d = v_0 t + \tfrac{1}{2}at^2 \) | \( v^2 = v_0^2 + 2ad \) | \( d = \tfrac{1}{2}(v_0+v)t \)
\( g = 9.8 \text{ m/s}^2 \) | Range \( R = v_0^2\sin(2\theta)/g \) | \( c^2 = a^2+b^2 \)
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Question 1 — Average velocity
A jogger runs 800 m east in 4.0 min, then 400 m west in 2.0 min. What is the average velocity for the whole trip?
Answer: m/s [E]
Solution:
Net displacement: \( 800 - 400 = 400 \text{ m east} \)
Total time: \( 4.0 + 2.0 = 6.0 \text{ min} = 360 \text{ s} \)
\( v_\text{avg} = 400/360 = 1.11 \text{ m/s [E]} \)
Question 2 — Uniform acceleration
A car accelerates uniformly from \(8.0\) m/s to \(28\) m/s over a distance of \(120\) m. What is the acceleration?
A boat heads directly across a river at \(4.0\) m/s relative to the water. The current flows downstream at \(3.0\) m/s. What is the boat's resultant speed relative to the shore?
Answer: m/s
Solution:
Velocities are perpendicular: \( v = \sqrt{4^2+3^2} = \sqrt{25} = 5.0 \text{ m/s} \)
Question 9 — Concept (multiple choice)
A ball thrown straight up returns to the thrower's hand. At the highest point of its motion, which statement is correct?
Solution:
At the apex the ball is momentarily at rest (\( v=0 \)) but gravity continues to accelerate it downward at \( g = 9.8 \) m/s², which is what brings it back down.
Question 10 — Stopping distance
A car travelling at \(25\) m/s applies the brakes and decelerates uniformly at \(5.0\) m/s². How far does the car travel before stopping?