Write the four equations of motion for uniform acceleration.
State the condition that must hold for these equations to apply.
A car accelerates uniformly from 8.0 m s−1 to 20 m s−1 over 60 m. Find the acceleration and the time taken.
A car travelling at 30 m s−1 brakes at 6.0 m s−2. Taking the direction of motion as positive, find the stopping distance and the time to stop.
Explain why these equations cannot be used when the acceleration changes during the motion.
State what you would use instead to find the displacement when the acceleration varies.
Total: 26 marks. Original work by the TheLucidSTEM team. Written in the style of the papers; no past paper question is reproduced.
Answer key · full worked solutionsclick to reveal
v = u + a t; s = ½(u + v) t; s = u t + ½ a t²; v² = u² + 2 a s.
the acceleration must be constant (uniform) throughout the motion.
B1. v = u + a t = 5.0 + 2.0 × 6.0 = 17 m s−1.
B2. s = u t + ½ a t² = 0 + ½ × 4.0 × 3.0² = 18 m.
B3. s = ½(u + v) t = ½(6.0 + 14) × 10 = 100 m.
B4. v² = u² + 2 a s = 0 + 2 × 2.5 × 20 = 100, so v = 10 m s−1.
time is missing, so use v² = u² + 2 a s: 20² = 8.0² + 2 a (60), so 400 = 64 + 120 a, giving a = 2.8 m s−2. Then v = u + a t: 20 = 8.0 + 2.8 t, so t = 12 / 2.8 = 4.3 s.
with the direction of motion positive, a = −6.0 m s−2 and v = 0. Distance: 0 = 30² + 2(−6.0)s, so s = 900 / 12 = 75 m. Time: 0 = 30 + (−6.0)t, so t = 5.0 s.
the equations are derived from a velocity-time graph that is a straight line, which assumes a constant acceleration. If the acceleration changes, the line is no longer straight and the derivation fails.
use the area under the velocity-time graph to find the displacement, and the gradient of a tangent to find the instantaneous acceleration.