Find the total distance the particle travels from t=0 seconds to t=5 seconds.

Question

A particle moves along a line with a velocity v(t)=−t+3, measured in meters per second. Find the total distance the particle travels from t=0 seconds to t=5 seconds.

Answer 1

Total distance s = ∫₀⁵I- t + 3Idt = ∫₀³(- t +3)dt + ∫₃⁵(t - 3)dt = (- t²/2 + 3t)₀³ + (t²/2 - 3t)₃⁵ = (- 9/2 + 9) + (25/2 - 15) - (9/2 - 9) = 9 -2.5 = 6.5

Answer 2

v = dx/dt therefore dx = vdt and, integrating both sides, we get x = ∫v dt however this "x" is the displacement and may or may not match the total distance traveled. If the particle goes one direction then back to where it started then its total displacement is zero but the distance traveled is certainly not zero.

Notice that in this case the velocity for the first 3 seconds is going to be negative and the velocity after that will be positive. So let's break this up into two parts: 0 - 3 seconds (we'll call this x₁) and 3 - 5 seconds (we'll call this x₂):

The antiderivative of -t + 3 is -1/2t² + 3t + C because when we take the derivative of -1/2t² + 3t + C we get -t + 3. Since this is a definite integral, we can ignore the "C" in the antiderivative because it will just subtract out when we do the arithmetic.

x₁ = ₀∫³(-t + 3) dt = [-1/2t² + 3t]₀³ = [-1/2(3)² + 3(3)] - [-1/2(0)² + 3(0)] = [-9/2 + 9] - [0] = 4.5

x₂ = ₃∫⁵(-t + 3) dt = [-1/2t² + 3t]₃⁵ = [-1/2(5)² + 3(5)] - [-1/2(3)² + 3(3)] = [-25/2 + 15] - [-9/2 + 9] = 2.5 - 4.5 = -2 but since we want total distance, we can consider that we want the absolute value of -2 which is 2.

So the total distance traveled is 4.5 + 2 = 6.5 meters

Answer 3

Compute the integral: ∫₀⁵ (-t+3)dt