Question

An object moves in one dimensional motion with constant acceleration a = 3.2 m/s². At time t = 0 s, the object is at x₀ = 2.2 m and has an initial velocity of v₀ = 4.9 m/s.

How far will the object move before it achieves a velocity of v = 8.4 m/s?

Your answer should be accurate to the nearest 0.1 m.

Answer 1

$v_f=v_0+at$

$t=\frac {v_f-v_0}{a}=\frac {8.4-4.9}{3.2}\: s= 1.094\: s$

$x=x_0+v_0t+\frac {1}{2}at^2$

$x=2.2+4.9\cdot 1.094+\frac {1}{2}\cdot 3.2\cdot 1.094^2$

$x=9.476\: m$