Answer:
He traveled 22.5 meters during his acceleration
Explanation:
Jeremy is riding his bike down a straight road at 3 m/s when he
accelerates at a constant rate
⇒ His initial speed = 3 m/s
He accelerates at a constant rate for 5 seconds
⇒ The time of acceleration = 5 seconds
He reaches a speed of 6 m/s
⇒ His final speed = 6 m/s
We need to find how far he traveled during his acceleration
Lets find his constant acceleration at first
Acceleration is the rate of change in velocity
[tex]a=\frac{v-u}{t}[/tex], where v is the final velocity , u is the initial velocity
and t is the time of acceleration
⇒ [tex]a=\frac{6-3}{5}=0.6[/tex] m/sec²
Now lets find the distance using the rule [tex]s=ut+\frac{1}{2}at^{2}[/tex]
where s is the distance
⇒ [tex]s=3(5)+\frac{1}{2}(0.6)(5)^{2}[/tex]
⇒ s = 15 + 7.5 = 22.5
The distance is 22.5 meters
He traveled 22.5 meters during his acceleration
