Question

a car is speeding up and has an instantaneous speed of 1.17 m/s when a stopwatch reads 11.6 s. it has a constant acceleration of 2.36 m /s^2. what change in speed occurs between t=11.6 and t=13.0 s?

Answer 1

$v_2 = v_1 + a\Delta t = 1.17 \frac{\text{m}}{\text{s}} + 2.36 \frac{\text{m}}{\text{s}^2}(13.0-11.6)\text{ s} = 4.474\frac{\text{m}}{\text{s}}.$