How to find distance when time and acceleration are given

Finding distance when you know time and acceleration depends critically on whether the object started from rest or had an initial velocity. If starting from rest (u = 0), the formula simplifies to s = ½at², where you multiply half the acceleration by the square of the time. For a car accelerating at 3 m/s² for 6 seconds from a standstill, the distance is s = ½(3)(36) = 54 meters. This quadratic relationship means distance increases dramatically with time—doubling the time quadruples the distance when acceleration is constant.

When initial velocity exists, you must use the complete formula s = ut + ½at² to account for distance covered at the starting speed plus distance gained through acceleration. If that same car already traveled at 10 m/s before accelerating at 3 m/s² for 6 seconds, the calculation becomes s = 10(6) + ½(3)(36) = 60 + 54 = 114 meters. The first term (60 m) represents constant-velocity distance, while the second term (54 m) represents acceleration-added distance—together they total actual displacement.

Verify your answer by cross-checking with final velocity calculations. Using v = u + at, the final velocity is 10 + 18 = 28 m/s. The average velocity is (10 + 28) / 2 = 19 m/s, and distance equals average velocity times time: 19 × 6 = 114 meters ✓. This confirmation method catches algebraic errors and reinforces the interconnected nature of kinematic variables. Always specify whether your calculated distance represents straight-line displacement or total path length—these differ in non-linear motion.