Differentiation: Velocity and Acceleration!

Calculus Level 1

Given the position function of a particle $s(t)= 2t^3 - 15t^{2} + 36t - 22$ , where $t$ is greater than or equal to 0, find the acceleration function at time $t$ .

$s$ is measured in metres and $t$ is measured in seconds.

a(t) = 18t - 14

a(t) = 6t^{2} -30t + 36

a(t) = 12t - 30

1 solution

Denis Nadarević
Feb 13, 2016

If you've taken a basic calculus class, you probably would have learned basic derivative rules, especially when it comes to the application of velocity and acceleration. Given a position function, you can determine the velocity and acceleration function.

the derivative of the position function is the velocity function, which is: $v(t) = 6t^{2} - 30t + 36$

You are asked to determine the acceleration function. The derivative of the velocity function is the acceleration function, which is: $a(t) = 12t - 30$

Differentiation: Velocity and Acceleration!

1 solution

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