(5.2) - Higher-Order Derivatives Velocity and Acceleration
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(5.2) - Higher-Order Derivatives Velocity and Acceleration Ex.) Given y = f(x) = 3x 4 - 6x 3 y' = f'(x) = dy = dx y'' = f''(x) = d 2 y = dx 2 y''' = f'''(x) = d 3 y = dx 3 = d 4 y = dx 4 = d 5 y = dx 5 In general, the number of derivatives equals the degree + 1 MCV4U1
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MCV4U1. (5.2) - Higher-Order Derivatives Velocity and Acceleration. Ex.) Given y = f(x) = 3x 4 - 6x 3 y' = f'(x) = dy = dx y'' = f''(x) = d 2 y = dx 2 y''' = f'''(x) = d 3 y = dx 3 = d 4 y = dx 4 = d 5 y = - PowerPoint PPT Presentation
Transcript of (5.2) - Higher-Order Derivatives Velocity and Acceleration
(5.2) - Higher-Order Derivatives Velocity and Acceleration
Ex.) Given y = f(x) = 3x4 - 6x3
y' = f'(x) = dy = dx
y'' = f''(x) = d2y = dx2
y''' = f'''(x) = d3y = dx3
= d4y = dx4
= d5y = dx5
In general, the number of derivatives equals the degree + 1
MCV4U1
Velocity v(t)- the rate of change (derivative) of distance s(t) with respect to time
v(t) = s'(t)
Acceleration A(t)- the rate of change (derivative) of velocity v(t) with respect to the time (t)
A(t) = v'(t) = s''(t)
Speed = |v(t)|NOTE: speed cannot be measured in negative
numbers
If [a(t)] > 0 accelerationIf [a(t)] < 0 deceleration
Example #1A ball is thrown upwards so that its distance above the ground in metres after t seconds is s(t) = 24.5t - 4.9t2
a) Find v(t) and a(t)
v(t)=s'(t)=
a(t)=v'(t)=s''(t)=
b) Find the max height of the ball**v(t)= at the max**
Example #2The position function of a particle is given by s(t) = t3 + 2t2 + 2t (s in metres, t in seconds)
a) Determine v(t) and a(t), for any time t
b) What is the velocity and acceleration at 3 seconds?
c) Find the velocity when a(t) = 34 m/s2
d) When is acceleration negative?
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