Question

The height, in feet, of a particle from the ground is given by the function s(t)=2.5t^(2)+13t, where 0<=t<=16. Find the velocity of the particle at t=3.

Answer 1

The velocity of the particle at t=3 seconds is determined by differentiating the position function with respect to time and then substituting t=3 into the resulting velocity function, which yields 28 feet per second.

Step-by-step explanation:

The question asks to find the velocity of a particle at a specific time given its position function s(t)=2.5t2+13t. To find the velocity, we need to take the first derivative of the position function with respect to time. The derivative s'(t) will give us the velocity v(t) at any time t.

Step-by-step solution:

• First, differentiate the function s(t) with respect to t to find the velocity function: v(t) = s'(t) = d/dt [2.5t2+13t] = 5t + 13.
• Substitute t = 3 into the velocity function to find the velocity at that instant: v(3) = 5(3) + 13 = 15 + 13 = 28 feet per second.

Thus, the velocity of the particle at t = 3 seconds is 28 feet per second.