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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?

2 Answers

3 votes
5/4 is your answer buddy because you need divide and multiply
User SBurris
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5.7k points
3 votes

Answer:

a)4.4s

b)5.4s

Step-by-step explanation:

a) power =
(workdone)/(time)

for a body in motion

workdone =ΔK.E=
(1)/(2)mv₂² ₋

v₂=final velocity=17m/s

v₁=initial velocity=0m/s

m=mass of car=825kg

work done=
(1)/(2)×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 =
(1)/(2)mv₂² + mgh

woekdone=
(1)/(2)×825×17² + 825×9.8×3.2

=119212.5 + 25,872= 145,084.5‬J

time = workdone ÷ power

=145084.5 ÷ 26856

=5.4s

User Vincent Menzel
by
4.9k points