If a ball is thrown in the air with a velocity 34 ft/s, its height in feet t seconds later is given by y = 34t - 16t^2.
Find the average velocity for the time period beginning when t = 2 and lasting 0.1 second.
How do I do this? My hw is due in 29 mins and I don’t know what to do
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2 responses so far ↓
1 corsica_band // Sep 18, 2008
velocity is the first derivative.
y’=34-32t
now plug in 2 for t
y’=34-32(2)
=-2 ft/s
Meaning has just begun falling from when you threw it upward.
2 Scott H // Sep 18, 2008
Velocity is the time rate of change of position. So,
dy/dt = v = 34 - 32*t
v(2) = 34 - 32*(2) = -30 ft/s
v(2.1) = 34 - 32*(2.1) = -33.2 ft/s
average velocity = (v(2) + v(2.1))/2 = -31.6 ft/s
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