An object with a velocity of 4.0 m/s is accelerated at a rate of 1.2 m/s ^2 for 5.0 s . Determine the distance travelled during this period of acceleration in m.

Step-by-step solution

Using the second equation of motion: s = ut + 1 2 at^2 Substituting the given values, u = 4.0 m/s , a = 1.2 m/s ^2 , and t = 5.0 s : s = (4.0)(5.0) + 1 2 (1.2)(5.0)^2 s = 20.0 + (0.6)(25.0) s = 20.0 + 15.0 = 35 m