How do I solve this problem without integrals or derivatives?
Find the distance traveled in 14 seconds by an object that is moving with a velocity of v(t) = 11 + 6cos t feet per second.
A. 154.8204
B. 156.1704
C. 159.9436
I don't see any way, other than numerical approximations. The velocity is not a nice linear function, so only an integral can give an exact answer
∫[0,14] (11+6cosx) dx
How do I get the numerical approximations?
Riemann sums, trapezoidal method.
google numerical integration or quadrature
To solve this problem without using integrals or derivatives, you can use a numerical method called the Trapezoidal Rule. The Trapezoidal Rule approximates the area under a curve by dividing it into trapezoids.
Here are the steps to apply the Trapezoidal Rule to find the distance traveled:
1. Divide the time interval into smaller subintervals. In this case, we have a time interval of 14 seconds, so you can choose any number of subintervals. Let's use n subintervals for simplicity.
2. Determine the width of each subinterval. Since we have 14 seconds divided by n subintervals, the width of each subinterval is Δt = 14/n seconds.
3. Calculate the velocity at each subinterval. Start with t=0 as the initial time. For each subinterval, calculate the velocity using the given velocity function v(t) = 11 + 6cos(t). Substitute the value of t into the function to find the velocity at that particular time.
4. Add up the velocities from all the subintervals. The sum of all the velocities will give you an approximation of the total distance traveled.
5. Multiply the sum of velocities by the width of each subinterval (Δt). This step accounts for the time duration of each subinterval.
6. Add up the multiplied values from all the subintervals, and that will give you an approximation of the total distance traveled.
7. Round the result to the appropriate number of decimal places, as given in the answer choices.
Repeat steps 1-6 for different values of n (number of subintervals), and see which value of n gives the desired answer choice.