Question

7.38 Power Part A How long will it take an 992-kg car with a useful power output of 35.6 hp (1 hp 746 W) to reach a speed of 14.0 m/s, neglecting friction? Tries 0/10 Submit Answer Part B How long will this acceleration take if the car also climbs a 3.45-m high hill in the process? Tries 0/10 Submit Answer O 1998-2019 by Florida State University. All rights reserved The contents of this web site are for use by the students enrolled in this course only. Any distribution of the contents in whole or in parts and in any form, including posting on the web or in printed version, is a violation of the Copyright laws and the FSU honor codes.

Answer 1

Part A:

mass of car = 992 kg

power output = 35.6 * 746 = 26557.6 W

final velocity = 14 m/s

Now, the power is defined as the rate of doing work and work is the form of energy so the kinetic energy of the car at 14 m/s velocity,

W = (1/2)*M*V²

W = 0.5 * 992 * 14²

W = 97216 Joules

now Power,

P = W/t

where P is the power and W is the work done.

26557.6 = 97216/t

solving for time,

t = 3.66 seconds

It will take 3.66 seconds to reach a speed of 14 m/s.

Part B:

for finding acceleration,

using first equation of motion,

V = U + a*t

where V and U are the final and initial velocity velocities respectively.

using known values,

14 = 0 + 3.66*a

solving for acceleration,

a = 3.82 m/s²

Now the time taken by car to move 3.45 m hill,

using second equation of motion,

S = U*t + (0.5)*a*t²

using the known values,

3.45 = 14*t + 1.96*t²

solving for t,

t = 0.24 seconds

car will take 0.24 seconds to climb a 3.45 m high hill in the process.