Question

A bike first accelerates from 0.0m/s to 4.5m/s in 4.5 s, the continues at this constant speed for another 6.0 s. What is the total distance traveled by the bike?

Answer 1

Answer:

37.125 m

Explanation:

Using the equation of motion

s=ut+0.5at^{2} where s is distance, u is initial velocity, t is time and a is acceleration

Distance during acceleration

Acceleration, a=\frac {V_{final}-V_{initial}}{t} where V_{final} is final velocity and V_{initial} is initial velocity.

Substituting 0.0 m/s for initial velocity and 4.5 m/s for final velocity, acceleration will be

a=\frac {4.5 m/s-0 m/s}{4.5 s}=1 m/s^{2}

Then substituting u for 0 m/s, t for 4.5 s and a for 1 m/s^{2} into the equation of motion

s=0*4.5+ 0.5*1*4.5^{2}=0+10.125 =10.125 m

Distance at a constant speed

At a constant speed, there's no acceleration and since speed=distance/time then distance is speed*time

Distance=4.5 m/s*6 s=27 m

Total distance

Total=27+10.125=37.125 m