Question

Two bacteria are next to each other moving in the same direction. One moves with an initial velocity of 20 μm/s, accelerating at a rate of 5 μm/s2. The other starts with a velocity of 60 μm/s and is decelerating at a rate of 2 μm/s2. 1 μm=10-6 m.
(a) Find the time and position at which the bacteria meet again for an epic battle

I just started 1-D Kinematics, stuff like this just confuses me. I'd more so like the "how to do" rather than just the answer so I can still learn.

Answer 1

We want to find the motion equations for two bacteria, and use these to see when the two bacteria will meet again.

We will see that the bacteria meet again after 26.67 seconds.

Let's call bacteria number 1 at the one that moves with an initial velocity of 20 μm/s and that accelerates at 5 μm/s^2.

Remember that the acceleration is the rate of change of the velocity, then the velocity equation of this bacteria is:

$v_1(t) = (5 \mu m/s^2)*t + 20 \mu m/s$

The position equation is given by the integration of the above equation:

$p_1(t) = (1/2)* (5 \mu m/s^2)*t^2 + (20 \mu m/s)*t + p0$

Where p0 is the constant of integration, this is the initial position of the bacteria number one. We can define this equal to zero, because we can decide where the zero of our axis is.

$p_1(t) = (1/2)* (5 \mu m/s^2)*t^2 + (20 \mu m/s)*t$

Similarly, for the other bacteria we can write the equations:

$v_2(t) = (-2 \mu m/s^2)*t + 60 \mu m/s$

Again we integrate to get the position equation:

$p_2(t) = (1/2)* (-2 \mu m/s^2)*t^2 + (60 \mu m/s)*t$

a) We want to see when the bacteria meet again, then we just need to solve:

$p_1(t) = p_2(t)$

$(1/2)* (5 \mu m/s^2)*t^2 + (20 \mu m/s)*t = (1/2)* (-2 \mu m/s^2)*t^2 + (60 \mu m/s)*t\\\\(1/2)*(5 \mu m/s^2)*t^2 + (1/2)*(2 \mu m/s^2)*t^2 + (20 \mu m/s)*t - (60 \mu m/s)*t = 0$

We will get a quadratic equation, simplifying the above we get:

$(1.5 \mu m/s^2)*t^2 - (40 \mu m/s)*t = 0\\\\(1.5 \mu m/s^2)*t - (40 \mu m/s) = 0\\\\(1.5 \mu m/s^2)*t = (40 \mu m/s) \\\\t = (40 \mu m/s)/(1.5 \mu m/s^2) = 26.67s$

We can conclude that the bacteria will meet again after 26.67 seconds.

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