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(5.2) - Higher-Order Derivatives Velocity and Acceleration

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 =

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(5.2) - Higher-Order Derivatives Velocity and Acceleration

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  1. MCV4U1 (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

  2. v(t) = s'(t) A(t) = v'(t) = s''(t) Velocity v(t)- the rate of change (derivative) of distance s(t) with respect to time Acceleration A(t)-the rate of change (derivative) of velocity v(t) with respect to the time (t) Speed =|v(t)| NOTE: speed cannot be measured in negative numbers If [a(t)] > 0 acceleration If [a(t)] < 0 deceleration

  3. Example #1 A 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**

  4. Example #2 The 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? Page 185-187 #3, 4, 7-12, 14

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