Position,(distance) is the 2nd antiderivative of acceleration, so first you need to make an equation for acceleration, being 90=30+6t.
From there you would have to antiderivate the equation twice, using an intial value for velocity and position to solve for the C's that are created when taking the antiderivative of something, so without that information I can't help you. When you get that you would need to simply plug in 10 for your t value and there is your answer
The maker of an automobile advertises that it takes 10 seconds to accelerate from 30 kilometers per hour to 90 kilometers per hour. Assuming constant acceleration, compute the following
distance traveled in 10 seconds
2 answers
a = (90-30)/10 = 6 km/hr-s
s = 1/2 at^2 = 3 km/hr-s * (10s)^2
= 300 km-s/hr
= 300 km-s/hr * 1hr/3600s
= 300/3600 km
= 1/12 km
or, since you want to use calculus,
a = 6km/hr-s
= 6km/hr-s * 1hr/3600s
= 1/600 km/s^2
so, a(t) = 1/600
v(t) = 1/600 t
s(t) = 1/1200 t^2
s(10) = 100/1200 = 1/12 km
s = 1/2 at^2 = 3 km/hr-s * (10s)^2
= 300 km-s/hr
= 300 km-s/hr * 1hr/3600s
= 300/3600 km
= 1/12 km
or, since you want to use calculus,
a = 6km/hr-s
= 6km/hr-s * 1hr/3600s
= 1/600 km/s^2
so, a(t) = 1/600
v(t) = 1/600 t
s(t) = 1/1200 t^2
s(10) = 100/1200 = 1/12 km
1 answer