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How long (in s) will it take an 825 kg car with a useful power output of 36.0 hp (1 hp = 746 W) to reach a speed of 17.0 m/s, neglecting friction? (Assume the car starts from rest.) s (b) How long (in s) will this acceleration take if the car also climbs a 3.20 m-high hill in the process?

Answer :

jeff8133
5/4 is your answer buddy because you need divide and multiply
adetoseen

Answer:

a)4.4s

b)5.4s

Explanation:

a) power = [tex]\frac{workdone}{time}[/tex]

for a body in motion

workdone =ΔK.E= [tex]\frac{1}{2}[/tex]mv₂² ₋

v₂=final velocity=17m/s

v₁=initial velocity=0m/s

m=mass of car=825kg

work done=[tex]\frac{1}{2}[/tex]×825×17² -  

work done=119212.5J

from question,

power=36×746=26856W

26856=119212.5÷time

time=119212.5÷26856

time=4.4s

b) the car is experiencing two forms of energy

workdone =  [tex]\frac{1}{2}[/tex]mv₂² + mgh

woekdone=  [tex]\frac{1}{2}[/tex]×825×17² + 825×9.8×3.2

=119212.5 + 25,872= 145,084.5‬J

time = workdone ÷ power

=145084.5 ÷ 26856

=5.4s

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