Question
A distance ‘s’ of an object in metres is given by the following equation:
s=7t^2 * e^-0.4t
Use differential calculus to find the equations for velocity and acceleration and hence find:
Acceleration when t = 3 seconds
The maximum value of s after t = 0
s=7t^2 * e^-0.4t
Use differential calculus to find the equations for velocity and acceleration and hence find:
Acceleration when t = 3 seconds
The maximum value of s after t = 0
Answers
oobleck
use the chain rule and the product rule
s = 7t^2 e^(-0.4t)
v = ds/dt = e^(-0.4t) (14t - 2.8t^2)
a = dv/dt = e^(-0.4t)(14 - 11.2t + 1.12t^2)
s = 7t^2 e^(-0.4t)
v = ds/dt = e^(-0.4t) (14t - 2.8t^2)
a = dv/dt = e^(-0.4t)(14 - 11.2t + 1.12t^2)