How to find the time when speed, distance, and acceleration are known

When you know initial speed (u), distance (s), and acceleration (a), finding time requires using the kinematic equation s = ut + ½at² and solving this quadratic equation for t. Rearranging into standard quadratic form gives ½at² + ut - s = 0, which you solve using the quadratic formula: t = [-u ± √(u² + 2as)] / a. Since time cannot be negative in these contexts, select the positive root from the ± symbol.

Let's work through a practical example: a car traveling at 15 m/s accelerates at 2 m/s² over a distance of 200 meters—how long did this take? Plug into the formula: t = [-15 + √(225 + 800)] / 2 = [-15 + √1025] / 2 = [-15 + 32.02] / 2 = 8.51 seconds. You can verify this by substituting back: s = 15(8.51) + ½(2)(8.51²) = 127.65 + 72.42 = 200.07 meters ✓. The small rounding difference confirms the answer's accuracy.

An alternative approach uses v² = u² + 2as to first find final velocity, then uses v = u + at to find time, avoiding quadratic equations altogether. Using the same example: v² = 225 + 800 = 1025, so v = 32.02 m/s. Then t = (v-u)/a = (32.02-15)/2 = 8.51 seconds. This two-step method often feels more intuitive and reduces algebraic errors, though both approaches yield identical results. Choose based on your comfort with quadratic equations versus your confidence in multi-step calculations.