Choose the Right Equation for the Data You Are Given

Stopping distance from speed u with deceleration a is u²/(2a), so it scales as u²: cars at u and 4u stop in distances in the ratio 1 : 16. The distance covered in the nth second is sₙ = u + a (2n - 1)/2, so a body from rest covering 120 cm in the 6th second has a x 11/2 = 1.2 m, giving a = 0.218 m/s². From rest, distances in equal successive time intervals are in the ratio 1 : 3 : 5 : 7.

For a train or car that speeds up and then slows down to rest, the maximum speed is common to both stages. A train that accelerates at α for time t1 and then retards at β for time t2 has α t1 = β t2, so t1/t2 = β/α, and since the average speed of each stage is the same (half the maximum speed), x1/x2 = t1/t2 = β/α. A car that accelerates at f through a distance s, runs at constant speed for time t, then decelerates at f/2 to rest covers v²/(f) = 2s while stopping, so the total distance is s + vt + 2s = 5s gives v t = 2s, and with v² = 2 f s this yields s = (1/2) f t².

A particle that covers half the distance at 3 m/s and the other half in two equal time intervals at 4.5 and 7.5 m/s covers the second half at an average of (4.5 + 7.5)/2 = 6 m/s. For distance D, the time is D/6 + D/12 = D/4, so the average speed is 4 m/s. Never average the speeds directly unless equal times are spent at each speed.

In uniformly accelerated motion the velocity-time graph is a straight line, the displacement-time graph is a parabola and the acceleration-time graph is a horizontal line. Any graph that does not have these shapes is not uniform acceleration, whatever the numbers are.