Canada’s Wonderland has a roller coaster named ‘Leviathan’. It boasts a maximum height of 93 m, 80⁰ drop, speeds of 148 kms and is 1673m long. At its maximum height Mr. Currie drops his gum from his mouth. The height of the gum can be given by the mathematical model:
h = 93 - 4.9t2, where t is in seconds and h is measured in metres.
a) Find the average velocity of the gum on the intervals 2<t<3 and 2<t<2.1
b) Find the instantaneous velocity when t = 1.5.
I got dizzy watching the video of the ride. Never mind the ride!
(a) average velocity
- t∈(2,3)
Average velocity = v(final)-v(initial)
=(93-4.9(3²)) - (93-4.9(2²)
=4.9(9-4)
=24.5 m/s
- t∈(2,2.1)
similar to above.
(b)
h(t)=93-4.9t²
instantaneous velocity at time t
v(t)=h'(t)
=-9.8t
At t=1.5
v(t)=h'(t)=-9.8*1.5=14.7 m/s
To find the average velocity of the gum on the given intervals, we need to calculate the change in height and divide it by the change in time.
a) Interval 2 < t < 3:
To find the average velocity on this interval, we need to calculate the changes in height and time.
Change in height (Δh) = h(3) - h(2)
= (93 - 4.9 * (3^2)) - (93 - 4.9 * (2^2))
= 93 - 44.1 - 93 + 19.6
= -24.5 meters
Change in time (Δt) = 3 - 2
= 1 second
Average velocity = Δh/Δt
= -24.5 / 1
= -24.5 m/s
Interval 2 < t < 2.1:
Similarly, we calculate the change in height and time for this interval.
Change in height (Δh) = h(2.1) - h(2)
= (93 - 4.9 * (2.1^2)) - (93 - 4.9 * (2^2))
= 93 - 46.5589 - 93 + 19.6
= -19.9589 meters
Change in time (Δt) = 2.1 - 2
= 0.1 second
Average velocity = Δh/Δt
= -19.9589 / 0.1
= -199.589 m/s
b) To find the instantaneous velocity when t = 1.5, we need to find the derivative of the height function with respect to time.
Given height function: h = 93 - 4.9t^2
Taking the derivative of both sides with respect to time (t):
dh/dt = d(93 - 4.9t^2)/dt
= -9.8t
Now substitute t = 1.5 into this derivative:
dh/dt = -9.8 * 1.5
= -14.7 m/s
Therefore, the instantaneous velocity when t = 1.5 is -14.7 m/s.