Kinematics practice questions.

A projectile is launched horizontally at 15 m s⁻¹ from the top of a cliff and lands 4.0 s later. Ignoring air resistance, the horizontal distance travelled is

• 3.75 m
• 60 m
• 78 m
• 240 m

Answer: 60 m

Horizontal velocity is constant, so x = v_x t = 15 × 4.0 = 60 m. The vertical motion does not affect the horizontal distance.

A car's velocity increases uniformly from 0 to 12 m s⁻¹ in 6.0 s, then remains constant at 12 m s⁻¹ for a further 4.0 s. Determine the total distance travelled.

• 48 m
• 120 m
• 84 m
• 36 m

Answer: 84 m

Distance is the area under the velocity–time graph: ½ × 12 × 6.0 = 36 m during acceleration, plus 12 × 4.0 = 48 m at constant velocity, giving 84 m.

A stone is launched at an angle above the horizontal. Air resistance is negligible. Which statement describes the stone at the highest point of its path?

• Its velocity is horizontal and its acceleration is directed vertically downwards
• Its velocity and acceleration are both zero
• Its velocity is momentarily zero and its acceleration is directed vertically downwards
• Its velocity is horizontal and its acceleration is zero

Answer: Its velocity is horizontal and its acceleration is directed vertically downwards

At the top only the vertical component of velocity is zero, so the velocity is purely horizontal, while the acceleration remains g directed downwards throughout the flight.

A train travelling at 30 m s⁻¹ brakes with a uniform deceleration of 2.5 m s⁻². Calculate the distance the train travels before coming to rest.

• 90 m
• 180 m
• 12 m
• 360 m

Answer: 180 m

Using v² = u² + 2as with v = 0: s = u²/(2a) = 30²/(2 × 2.5) = 180 m. Forgetting the factor 2 gives the common wrong answer 360 m.

An athlete runs exactly one complete lap of a 400 m closed track. Which row gives the distance travelled and the magnitude of the displacement?

• Distance 0, displacement 400 m
• Distance 400 m, displacement 400 m
• Distance 0, displacement 0
• Distance 400 m, displacement 0

Answer: Distance 400 m, displacement 0

Distance is the total path length (400 m), while displacement is the straight-line change in position, which is zero because the athlete finishes where they started.

A stone is thrown vertically upwards at 25 m s⁻¹ from the edge of a cliff and lands on a beach 45 m below the point of release. A student writes: “Taking upwards as positive: u = +25 m s⁻¹, a = −10 m s⁻², s = +45 m, so 45 = 25t − 5t².” Identify the error in this working.

• The acceleration becomes +10 m s⁻² once the stone is on the way down.
• The initial velocity should be −25 m s⁻¹, because the throw opposes the gravitational field.
• The displacement should be −45 m, because the stone lands below the release point.
• The final velocity, not the initial velocity, belongs in the term 25t.

Answer: The displacement should be −45 m, because the stone lands below the release point.

Choosing a positive direction fixes the sign of every vector for the whole flight: a stays −10 m s⁻² while the stone rises and while it falls, and a landing point below the release point gives a negative displacement. With s = +45 m the quadratic has no real root at all, which is the tell-tale sign that a sign, not the algebra, is wrong.