Asked by Caleb
The work W0 accelerates a car from 0 to 20 km/h. How much work (in terms of W0) is required to accelerate the car from 20 km/h to 100 km/h?
Answers
Answered by
Damon
Let's assume the acceleration, a , is constant
then the force is constant
F = m a
work = integral of force dx from low speed to high speed
which is force * (Xend - Xbegin) because F is constant
so we need to know x begin and x end
let x begin be 0
then
d^2x/dt^2 = a
dx/dt = v = Vi + a t
20 = 0 + a t
so t = 20/a
x = 0 + Vi t + (1/2) a t^2
Xend = (1/2) a (400/a^2) = 200/a
so
Wo = m a ( 200/a) = 200 m
now the second part
integrate the same thing but Vi = 20 not 0
F = m a still so
W = m a [ Xend - Xbegin] again
let x begin = 0 again
Vi = 20
v = 20 + a t
40 = 20 + a t
t = 20/a same time for the second part
but now our average speed is 60
Xend = 1200/a
so
W = m a (1200/a) = m (1200)
W/Wo = 1200/200 = 6
W = 6 Wo
by the way that means we need 6 times the power because the time is the same
then the force is constant
F = m a
work = integral of force dx from low speed to high speed
which is force * (Xend - Xbegin) because F is constant
so we need to know x begin and x end
let x begin be 0
then
d^2x/dt^2 = a
dx/dt = v = Vi + a t
20 = 0 + a t
so t = 20/a
x = 0 + Vi t + (1/2) a t^2
Xend = (1/2) a (400/a^2) = 200/a
so
Wo = m a ( 200/a) = 200 m
now the second part
integrate the same thing but Vi = 20 not 0
F = m a still so
W = m a [ Xend - Xbegin] again
let x begin = 0 again
Vi = 20
v = 20 + a t
40 = 20 + a t
t = 20/a same time for the second part
but now our average speed is 60
Xend = 1200/a
so
W = m a (1200/a) = m (1200)
W/Wo = 1200/200 = 6
W = 6 Wo
by the way that means we need 6 times the power because the time is the same
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