Question

A cyclist is initially traveling at 12 m/s. He travels 32 m for the next 4 seconds. Determine (a) Its acceleration (b) its speed after 4 seconds

Answer 1

ANSWER

• a = -2m/s²

,

• v = 4m/s

EXPLANATION

Given:

• The initial speed of the cyclist, v₀ = 12 m/s

,

• The distance traveled, x = 32m

,

• The time interval, t = 4s,

Unknown

• The acceleration, a

,

• The speed after 4s, v

The distance the cyclist travels, assuming he is traveling with constant acceleration is,

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

We have to solve this equation for a. Subtract v₀t from both sides,

$x-v_ot=\frac{1}{2}at^2$

And divide both sides by 1/2t²,

$a=\frac{2(x-v_ot)}{t^2}$

Replace with the given values,

$a=\frac{2(32m-12m/s\cdot4s)}{4^2s^2}=\frac{2(32m-48m)}{16s^2}=\frac{2\cdot(-16)m}{16s^2}=-2m/s^2$

Hence, the cyclist's acceleration is -2m/s².

The speed is,

$v=v_o+at=12m/s-2m/s^2\cdot4s=12m/s-8m/s=4m/s$

Hence, the final speed is 4m/s.