Question

A car traveling due east with an initial velocity of 10 m/s accelerates for 6 seconds at a constant rate of 4 m/s2. i.What is its velocity at the end of this time? ii.How far does it travel during this time?

Answer 1

From the kinematic equaiton of the motion with constant acceleration, the velocity after time $t = 6s$ is:

$$v = v_0 + at$$

where $v_0 = 10m/s$ is the initial velocity, $a = 4m/s^2$ is the acceleration. Thus, obtain:

$$v = 10m/s + 4m/s^2\cdot 6s = 34m/s$$

From another kinematic equation, the distance travelled in the same time is:

$$d = v_0t + \dfrac{at^2}{2}\\ d = 10m/s\cdot 6s + \dfrac{4m/s^2\cdot (6s)^2}{2} =132m$$

Answer. 34 m/s and 132 m.