(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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