We have s(t) = 12at2 + v0t + s0 = −16t2 + h. Therefore, the velocity isv(t) = s'(t) = .
Question
We have s(t) = 12at2 + v0t + s0 = −16t2 + h. Therefore, the velocity isv(t) = s'(t) = .
Solution
The velocity v(t) is the derivative of the position function s(t). In this case, s(t) = -16t^2 + h.
To find the derivative, we apply the power rule, which states that the derivative of t^n is n*t^(n-1).
Step 1: Identify the power of t in each term. In -16t^2, the power of t is 2. In h, the power of t is 0 (since h is a constant and can be thought of as h*t^0).
Step 2: Apply the power rule to each term. The derivative of -16t^2 is 2*-16t^(2-1) = -32t. The derivative of h is 0t^(0-1) = 0.
Step 3: Combine the derivatives of each term to get the derivative of the entire function. In this case, v(t) = s'(t) = -32t + 0 = -32t.
So, the velocity function v(t) = -32t.
Similar Questions
A stone was dropped off a cliff and hit the ground with a speed of 152 ft/s. What is the height of the cliff? (Use 32 ft/s2 for the acceleration due to gravity.)Part 1 of 4We know that s(t) = 12at2 + v0t + s0. In this situation, we have a = ft/s2, v0 = ft/s, and s0 = h, where h is the height of the cliff (in feet) that we wish to find.
The acceleration 'a' in m/s2 of a particle is given by a = 3t2 + 2t + 2 where t is the time. If the particle starts out with a velocity, u = 2 m/s at t = 0, then the velocity at the end of 2 seconds will be :12 m/s18 m/s27 m/s36 m/s
The acceleration function (in m/s2) and the initial velocity v(0) are given for a particle moving along a line.a(t) = 2t + 4, v(0) = −32, 0 ≤ t ≤ 6Exercise (a)Find the velocity at time t.Step 1The velocity function is the antiderivative of the acceleration.v(t) = (2t + 4) dt = $$t2+4t + CStep 2We must determine the value of C. We know that v(0) = −32.Substituting 0 into our antiderivative gives −32 = v(0) = 3 + C. Therefore, C = . Exercise (b)Find the distance traveled during the given time interval.Step 1The velocity function is v(t) = t2 + 4t − 32, and so the distance traveled in the time interval 0 ≤ t ≤ 6 is given by 6|t2 + 4t − 32| dt0.Remembering that |z| = z ≥ 0 z < 0, we must determine where v(t) = t2 + 4t − 32 is positive or negative.v(t) can be factored as t2 + 4t − 32 = t + t − .
The acceleration function (in m/s2) and the initial velocity v(0) are given for a particle moving along a line.a(t) = 2t + 2, v(0) = −15, 0 ≤ t ≤ 5(a) Find the velocity at time t.v(t) = m/s(b) Find the distance traveled during the given time interval.
We know that s(t) = 12at2 + v0t + s0. In this situation, we have a = ft/s2, v0 = ft/s, and s0 = h, where h is the height of the cliff (in feet) that we wish to find.
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.