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Iwo cars drive on a straight highway. At time $t=0,$ car 1 passesmile marker 0 traveling due east with a speed of 20.0 $\mathrm{m} / \mathrm{s} .$ At thesame time, car 2 is 1.0 km east of mile marker 0 traveling at 30.0 $\mathrm{m} / \mathrm{s}$due west. Car 1 is speeding up with an acceleration of magnitude$2.5 \mathrm{m} / \mathrm{s}^{2},$ and $\operatorname{car} 2$ is slowing down with an acceleration of mag-nitude 3.2 $\mathrm{m} / \mathrm{s}^{2}$ . (a) Write $x$ -versus-t equations of motion for bothcars, taking east as the positive direction. (b) At what time do thecars pass next to one another?

(a) Car $1 : x=20 t+1.25 t^{2} \mathrm{m}$Car $2 : x=1000-30 t+1.6 t^{2} \mathrm{m}$(b) 24.04 $\mathrm{sec}$

Physics 101 Mechanics

Chapter 2

One-Dimensional Kinematics

Motion Along a Straight Line

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okay. In this question, we need to write the equations of motion of two cars. Uh, let's see. First of all, we're going to define East us positive in the West US. Negative. So we know that the first car that's called Car One has initial position off zero because he passes to admire Mark zero. He's traveling due east, which means his speed is positive. So his initial speed off 20 needles per second and the car one has ah acceleration with magnitude 2.5 acceleration those two point five meters per second square and the car one is speeding up, which means the acceleration is in the same direction off the velocity. Okay, so what about a car? Two. The car too. We have initial position off one because he's one my own east. Let's see what else uh, occur to has velocity of 30 meters per second, initial velocity of dirty meters per second and a deceleration off 3.2 meters per second scrap. Since this is a deceleration, we're going to put a minor sign in front of the deceleration. So the questions of motions of both cars is going to be to the car one going to be ex. It grows. Drank tea tree trying T D plus 2.5, divided by two. T square inferred a car too. We're going to have acts E close one plus 30 t. Uh, Let's see. Eternity minus 3.2. Divided by two T Square. This is a D Square. Okay, so this is the both accretions of motions for both the cars. So the second I think we need to discover it And what moment they go into, they're going to intercept one another. So we just need to the moment when they intercept one another is when the final positions of the car one is equal to final positions of the car too. So we just need to match the both equations here. This is going to be 20 d plus 2.5, divided by two D square, 2.5 diversified to be scrapped equals to one. Tell us. Dirty T minus 3.2, divided by two T square. Okay, so let's rearrange this. Go. It's and have to minus two points. 80 five t square plus 10 t close. One equals zero. So this is a contract a creation, and we know how to solve quadratic creations. The solution for this contract creation in particular, is going to be let's see, 24.4 So did time going to be 24.4 seconds. And that's the answer. Thanks for watching.

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