Calculus in physics

Question

If the current position of the object at time t is s(t), then the position at time h later is s(t+h). The average velocity (speed) during that additional time h is (s(t+h)−s(t))h . If we want to analyze the instantaneous velocity at time t, this can be made into a mathematical model by taking the limit as h→0, i.e. the derivative s′(t). Use this function in the model below for the velocity function v(t).

 

The acceleration is the rate of change of velocity, so using the same logic, the acceleration function a(t) can be modeled with the derivative of the velocity function, or the second derivative of the position function a(t)=v′(t)=s′′(t).

 

Problem Set question:

 

A particle moves according to the position function s(t)=e^8tsin(5t).

 

Enclose arguments of functions in parentheses. For example, sin(2t).

 

(a) Find the velocity function.

v(t)=

(b) Find the acceleration function.

a(t)=

Answer 1

Hi Sabrina,

To find the v(t) function, we must take the derivative of s(t). This involves the product rule, as shown below:

v(t) = d/dt[e^8tsin(5t)] = sin(5t) * d/dt [e^8t] + e^8t * d/dt [sin(5t)] = 8e^8tsin(5t) + 5e^8tcos(5t)

= e^8t(8sin(5t) + 5cos(5t)).

The acceleration a(t) is the derivative of v(t), so

a(t) = v'(t) = d/dt [e^8t(8sin(5t) + 5cos(5t))] = e^8t * d/dt [8sin(5t) + 5cos(5t)] + (8sin(5t) + 5cos(5t)) * d/dt [e^8t] = e^8t(40cos(5t) - 25sin(5t)) + 8e^8t(8sin(5t) + 5cos(5t)) = e^8t(40cos(5t) - 25sin(5t)) + e^8t(64sin(5t) + 40cos(5t))

= e^8t(80cos(5t) + 39sin(5t)).

Hope this helps!

Akshat Y.