A car starts from rest and accelerates to 20 m/s, with constant acceleration, in the span of 10 seconds. It continues for 30 seconds and then decelerates back to rest. How far did the car travel?
We can break this problem in to multiple parts to solve it: acceleration, constant velocity, and deceleration. In all cases, we will use one of Newton's equations of motion to solve it. Constant velocity: d = v*t 20 m/s * 30 s = 600 m Acceleration: First we find the acceleration. a = (v2 - v1)/t = (20 m/s - 0)/10s = 2m/s/s Now we can find the distance traveled d = v_1 * t + .5* a * t^2 = 0*10s + .5 * 2m/s/s * 10*10 = 100 m Deceleration: First we find the acceleration. a = (v2 - v1)/t = (0 - 20m/s)/10s = -2m/s/s Now we can find the distance traveled d = v_2 * t + .5* a * t^2 = 20*10 - .5 * 2m/s/s * 10*10 = 100 m The total distance traveled would be 800 m.
Integrate the following expression: xe^x
We have to use integration by parts to integrate this expression. ∫ u*dv = u*v - ∫ v*du ∫ u*dv = ∫ xe^x dx u = x du=dx dv = e^x dx v = e^x ∫ xe^x dx = x*e^x - ∫ e^x dx = x*e^x - e^x
Find the first and second derivative of the following expression: 3x*cos(x)
Using the chain rule with x as one part and cos(x) as the other First: 3cos(x) - 3x*sin(x) Second: -3x*cos(x) - 6sin(x)