Question

A plane travelling at 52 m/s [W] down a runway begins accelerating uniformly at 2.8 m/s2 [W]. 

(a) What is the plane’s velocity after 5.0 s? 

(b) How far has it travelled during this 5.0 s interval?

Answer 1

a) The velocity is given by the kinematic formula:

$$v = v_0+at\\ v = 52m/s + 2.8m/s^2\cdot 5s = 66m/s$$

b) The distance is the following:

$$d = v_0t+\dfrac{at^2}{2}\\ d = 52m/s\cdot 5s + \dfrac{2.8m/s^2\cdot (5s)^2}{2} = 295m$$

Answer. a) 66m/s, b) 295m.