(III) Mary and Sally are in a foot race (Fig. 2–46). When Mary...

Question

(III) Mary and Sally are in a foot race (Fig. 2–46). When Mary is 22 m from the finish line, she has a speed of 4.0 m/s and is 5.0 m behind Sally, who has a speed of 5.0 m/s. Sally thinks she has an easy win and so, during the remaining portion of the race, slows down at a constant rate of 0.40 m/s² to the finish line. What constant acceleration does Mary now need during the remaining portion of the race, if she wishes to cross the finish line side-by-side with Sally?

Accepted Answer

Hey, everyone in this problem. Car P and KQ are moving towards the cabin. They both have different speeds. When car P is 98 m from the cabin, it's cruising at 45 m per second and is 22 m behind car Q which has a speed of 55 m per second. All right. So we're gonna come back to the rest of the question in just a second. But let's draw out what we've been given so far. So we have this cabin, this is our cabin and we have two cars, car P and car Q and car P. He's gonna be driving along and we're told that when it is 98 m from the cabin. And so the distance here is 98 m, it's going to be moving at 45 m per second. Ok. So this car is moving at 45 m per second. Now, we have a second car, car Q and it's in front of car P and we're told car P is behind KQ. So that means that KQ is in front of car P. So KQ and it is 22 m in front. So we have this distance of 22 m and Khaki is moving at 55 m per second. Ok. So this is the situation we have right now. Going back to the question, we're told that due to a mechanical issue, car Q is going to decelerate at a constant rate of 3.2 m per second squared for the rest of the way. And we're asked to determine the constant acceleration that car P needs to maintain in order to tie with car Q right at the end. Ok. So to finish at the cabin at the same time, so we have four answer choices in this problem all in meters per second squared. Option A 59 option B 32 option C 18 and option D 44. So let's think about car Q first, ok? We're trying to figure out the acceleration. Car P needs to reach the cabin at the same time as car Q, but let's determine how long it takes car Q to reach the cabin first. So we know what we're dealing with. Ok. So we are gonna start with car Q. We're gonna do that in ball. So car Q let's think about the variables we have, we know that its initial velocity is 55 m per second. Ok? We're gonna say that the direction to the right is positive. We don't know its final velocity. We know that it has a mechanical issue and is going to be decelerating 3.2 m per second squared. So our acceleration is going to be negative 3.2 meters per second squared. Pay that negative because it's a deceleration in delta X and the change in position the displacement. OK. What do we have? Well, the distance between car P in the cabin is 98 m, the distance between KQ and car P is 22 m. And so the distance that KQ has to go to reach the cabin is gonna be that 98 m minus 22 m and it's 22 m closer. And so that gives us 76 m. And what we're interested in here is how long this takes. So we have three known values V, not A and delta X, we want to find the time T. So we're gonna choose a kinematic or U AM equation with those four variables. And that equation is gonna be delta X is equal to V not multiplied by T plus one half multiplied by a multiplied by T squared substituting in our values 76 m is equal to 55 m per second, multiplied by the time T plus one half multiplied by negative 3.2 m per second squared multiplied by T squared. Now, looking at this equation, we can see that we have a quadratic equation in T and we have a T squared term A T term and then a constant term. So in order to solve this we wanna get everything to one side. We wanna have that equal to zero on the other side. And then we can use our quadratic equation. So let's simplify on the right hand side. First, we have 76 m is equal to 55 m per second, multiplied by T minus 1.6 m per second squared multiplied by T squared. And we're gonna move everything to the left hand side and we'll get that zero is equal to 1.6 m per second squared, multiplied by T squared minus 55 m per second, multiplied by T plus 76 m. So if we're thinking about our quadratic equation, our A value is gonna be the coefficient on T squared, 1.6. The B value is gonna be the coefficient on T negative 55. And our C value is gonna be that constant at the end of 76. So now let's use our quadratic equation. Bind this value for T. So we have T is going to be equal to negative B plus or minus the square root of B squared minus four AC, all divided by two A. And we can substitute in that A B and C value that we've written out. So T is going to be equal to 55 m per second plus or minus the square root of negative 55 m per second, all squared minus four multiplied by 1.6 m per second squared. Multiplied by 76 m. And all of this is divided by two, multiplied by 1.6 m per second squared. All right. So this looks messy, but we can simplify this down. And when we do, we're gonna get two T values. OK? We get one. When we add in the numerator, we get a, a separate one when we subtract in the numerator and we get that T is either equal to 1.44 234 seconds or 32.932 663 seconds. Ok? And what we're gonna do is we're just gonna go ahead and take that first T value and that first time that they reach the cabin together at the same time, we're gonna take that first value. Ok? All right. And now we're gonna switch over to looking at car P. Remember we're trying to find the constant acceleration that car P needs in order to reach the cabin at the same time as car key. We now know how long it's gonna take for car key to reach the cabin. So we have a target for KP uh switching over to car P and we're gonna do very similar process. We're gonna write out all of the information we know about car P. We know that the initial velocity is 45 m per second. We don't know the final velocity we're looking for the acceleration. We know that that car has to travel 98 m 98 m from the cabin. And it's going to take a time of 1.44234 seconds. And we want it to take that long so that they reach there at the same time. So this time, our three known values are VNO delta X and T, the thing we're looking for is the acceleration A. So we're gonna choose the same equation. OK? We're gonna exclude VF because we don't have it and that's not what we're looking for. So again, choosing the same equation, but we have different known variables this time. So for car P delta X is equal to V not multiplied by T plus one half, multiplied by a multiplied by T squared. Substituting in our values 98 m is equal to 45 m per second, multiplied by 1.44234 seconds, one half multiplied by the acceleration. A multiplied by 1.44234 seconds squared. Now let's go ahead and simplify on the right hand side. So we get 98 m is equal to 64 0.90 by 3 m plus a multiplied by 1.0402 2nd squared. We're gonna subtract 64.9 m from both sides. And we're left with a multiplied by 1.04 02 seconds squared is equal to 33.09 or 7 m. And finally dividing by our 1.0402 seconds squared. We get that the acceleration is going to be equal to about 31 0.816 m per second squared. Ok. And so in order to reach the cabin at the same time, car p must accelerate with an acceleration of about 31.8 m per second squared. Comparing this to our answer choices, we can round to the nearest meter per second squared and we get that the correct answer is going to be B 32 m per second squared. Thanks everyone for watching. I hope this video helped see you in the next one.

Struggling with Physics?

Watch next

Catch/Overtake Problems practice set