for all such average problems, use the difference quotient
(f(t+h) - f(t)) / h
If you think you're stuck, show some work.
The distance (in feet) of an object from a point is given by s(t)=3t2−5, where t is in seconds. Round your answer to two decimal places.
1. What is the average velocity of the object between t=1 and t=8.5?
2. By using smaller and smaller intervals around 1, estimate the instantaneous velocity at time t=1.
(tried several different ways and can't seem to get it right! Appreciate the help in advance!!)
2 answers
s(t) = 3t^2 − 5
I assume you mean location is s, not distance, because it is negative at t = 0
s(8.5) = 3* 8.5^2 - 5 = 211.75
s(1) = 3-5 = -2
distance moved in 7.5 seconds = 213.75
average velocity = 213.75 / 7.5 = 28.2 ft/s
for part b I will take the derivative
ds/dt = 6 t
at t = 1, velocity = 6 * 1 = 6 ft/s
your way"
t = 1
s = 3 * 1^2 - 5 = -2
at t = 1.1
s = 3 *1.1^2 - 5 = -1.37
av speed = (-1.37 - -2) / 0.1
= 0.63 /9.1 = 6.3
well the real answer is 6 so close
now try t = 1.01 :)
I assume you mean location is s, not distance, because it is negative at t = 0
s(8.5) = 3* 8.5^2 - 5 = 211.75
s(1) = 3-5 = -2
distance moved in 7.5 seconds = 213.75
average velocity = 213.75 / 7.5 = 28.2 ft/s
for part b I will take the derivative
ds/dt = 6 t
at t = 1, velocity = 6 * 1 = 6 ft/s
your way"
t = 1
s = 3 * 1^2 - 5 = -2
at t = 1.1
s = 3 *1.1^2 - 5 = -1.37
av speed = (-1.37 - -2) / 0.1
= 0.63 /9.1 = 6.3
well the real answer is 6 so close
now try t = 1.01 :)
1 answer