Particle's distance traveled

Two particles A and B start from rest at the origin s = 0 and move along a straight line such that a_A=(4t−2)ft/s² and a_B=(18t²−12)ft/s², where t is in seconds.

Determine the distance between them when t = 3 s. Express your answer using three significant figures and include the appropriate units.

You cab find the position equations by integrating the acceleration twice. The integral of acceleration over time is velocity, and the integral of velocity over time is position.

V_A = integral of a_A = 2t²-2t ft/s

V_B = integral of a_B = 6t³-12t ft/s

S_A = integral of V_A = 2/3 t³ - t² ft

S_B = integral of V_B = 3/2 t⁴ - 6t² ft

Now you plug in the 3 seconds into t for the positions to know where the particles where at 3, then you substract them to calculate the distance between them.

S_A(3) = 9 ft

S_B(3) = 67.5 ft

S_A(3) - S_B(3) = 58.5 ft

Determine the total distance particle A has traveled in t = 3 s. Express your answer using three significant figures and include the appropriate units.

To determine the total distance you don't just plug in 0 and 3 into the equation. This will only give you the displacement. Think about it, If a particle goes all the way to -100 in one second and comes back to 0 in two seconds, and you just plug in the 0 and the 3 and calculate the difference, you'll get 0 ft difference and think the particle didn't move, when in reality it traveled all the way to -100 and back, traveling a total of 200 ft. So, what you really need to ask your self is when did the particle change directions in this 3 second period? The answer is when the velocity is 0. So you need to look at the velocity functions and see when they equal 0.

V_A = 2t²-2t = 0 when t = 0 or t = 1

V_B = 6t³-12t = 0 when t = 0, t = -1.414 or t = +1.414

Then you plug in the values where the velocity equals 0 into the position formulas, you don't care about the values that fall outside the range that we are interested, for example, the -1.414 is outside the 3 second period so we ignore that one.

S_A(0) = 0 ft

S_A(1) = -1/3 ft (traveled 1/3 ft)

S_A(3) = 9 ft (traveled 9 1/3 ft)

So the total distance traveled is 9 2/3 ft or 9.67 ft

Determine the total distance particle B has traveled in t = 3 s. Express your answer using three significant figures and include the appropriate units.

S_B(0) = 0 ft

S_B(1.414) = -6 ft (traveled 6 ft)

S_B(3) = 67.5 ft (traveled 73.5 ft)

So the total distance traveled is 79.5 ft