A woman driving her car approaches an intersection and begins to accelerate just as the traffic light turns green. Her velocity is the function of time described by the equation:
v(t)=2t^2-8t+10
(A)
Use calculus to find her instantaneous acceleration at t=8.0seconds
(B)
Use calculus to find the area under the curve to see how far she traveled between t=0seconds and t=4.0seconds
Please walk me through all of your steps, instead of just suggesting a formula.
Thank you!
To find the instantaneous acceleration at a specific time, we need to differentiate the velocity function with respect to time. Let's start with part (A).
(A) Finding the instantaneous acceleration at t = 8.0 seconds:
Step 1: Differentiate the velocity function v(t) = 2t^2 - 8t + 10 with respect to time t.
Let's use the power rule for differentiation, which states that if f(t) = t^n, then the derivative of f(t) with respect to t is f'(t) = n*t^(n-1).
Differentiating our function v(t) = 2t^2 - 8t + 10:
v'(t) = (2*2t^1) - (8*1) + 0
v'(t) = 4t - 8
Step 2: Substitute t = 8.0 into v'(t) to find the acceleration at that specific time:
a(8.0) = 4*(8.0) - 8
a(8.0) = 32 - 8
a(8.0) = 24
Therefore, the woman's instantaneous acceleration at t = 8.0 seconds is 24 units per second squared.
Moving on to part (B):
(B) Finding the area under the curve to determine the distance traveled between t = 0 seconds and t = 4.0 seconds:
Step 1: Calculate the antiderivative of the velocity function v(t) = 2t^2 - 8t + 10 with respect to t.
Using the power rule of integration, the antiderivative of t^n with respect to t is (t^(n+1))/(n+1).
Regarding the constant term, integrating a constant 'c' just adds t to the equation.
So, the antiderivative of v(t) = 2t^2 - 8t + 10 will be:
s(t) = (2*t^(2+1))/(2+1) - (8*t^(1+1))/(1+1) + 10*t + c
s(t) = (2/3)*t^3 - 4t^2 + 10t + c
Step 2: Evaluate the integral using the limits of integration t = 0 seconds and t = 4.0 seconds to find the area under the curve, which represents the distance traveled.
s(4.0) - s(0) = (2/3)*(4.0^3) - 4*(4.0^2) + 10*(4.0) - [(2/3)*(0^3) - 4*(0^2) + 10*(0)]
s(4.0) - s(0) = (2/3)*(64.0) - 4*(16.0) + 40.0
s(4.0) - s(0) = (128/3) - 64 + 40
s(4.0) - s(0) = 42.7
Therefore, the woman traveled approximately 42.7 units of distance between t = 0 seconds and t = 4.0 seconds.
Sure! I'll walk you through the steps to find the answers to both parts of the question.
(A) To find the woman's instantaneous acceleration at t=8.0 seconds, you need to take the derivative of the velocity function with respect to time. The derivative of velocity with respect to time gives us acceleration.
Step 1: Start with the given velocity function: v(t) = 2t^2 - 8t + 10.
Step 2: Differentiate the velocity function with respect to time using the power rule of derivatives. The power rule states that d/dx(x^n) = n*x^(n-1).
So, differentiating the velocity function, we get:
a(t) = d/dt (2t^2 - 8t + 10)
= d/dt (2t^2) - d/dt (8t) + d/dt (10)
= 4t - 8.
Step 3: Now, substitute the given time value t=8.0 seconds into the acceleration function.
a(t=8.0) = 4(8.0) - 8
= 32 - 8
= 24.
Therefore, the woman's instantaneous acceleration at t=8.0 seconds is 24 (unit of acceleration).
(B) To find the area under the curve and determine how far the woman traveled between t=0 seconds and t=4.0 seconds, we need to calculate the definite integral of the velocity function over that time interval.
Step 1: Start with the given velocity function: v(t) = 2t^2 - 8t + 10.
Step 2: Integrate the velocity function with respect to time to find the displacement function. The definite integral represents the area under the curve. The integral of each term in the velocity function can be found using the power rule for integration.
∫[0 to 4] (2t^2 - 8t + 10) dt
= (2/3)t^3 - 4t^2 + 10t] [0 to 4]
= [(2/3)(4)^3 - 4(4)^2 + 10(4)] - [(2/3)(0)^3 - 0 + 10(0)]
= [(2/3)(64) - 4(16) + 10(4)] - [0]
= (128/3) - 64 + 40
= 42.67.
Therefore, the woman traveled approximately 42.67 units of distance between t=0 seconds and t=4.0 seconds.
I hope these step-by-step explanations have helped you find the answers to both parts of the problem. Let me know if you need any further assistance!