Determine the stopping distances for an automobile going a con...

Question

Determine the stopping distances for an automobile going a constant initial speed of 95 km/h in the +𝓍 direction, and human reaction time of 0.40 s:

(a) for an acceleration a = ― 2.5 m/s²"

Accepted Answer

Hey, everyone in this problem, a pickup truck is traveling along a straight path at a constant speed of 80 kilometers per hour given a human reaction time of 0.9 seconds and a deceleration rate of negative 2 m per second squared for as well. What is its stopping distance? We're given four answer choices all in meters. Option A 2.6 multiplied by 10. Option B 1.4 multiplied by 10 squared. Option C 5.8 multiplied by 10 and option D 3.2 multiplied by 10 squared. We have to think about this where as for a stopping distance, but the truck does not start decelerating immediately. We have this reaction time that we have to account for. And so the total stopping distance which we're gonna call delta X, that's what we're looking for. It's gonna be equal to delta XR the distance traveled or distance covered during the reaction time plus delta XD, the distance traveled during deceleration, right? So we have this total stopping distance that's made up of these two components. So let's start with the reaction phase. That delta XR value is what we want to calculate So during the reaction phase, the pickup truck is traveling at a constant speed of 80 kilometers per hour. And so VR is gonna be equal to 80 kilometers per hour. And because this is a constant speed or constant velocity, we only have one value. We don't need VN and VF we don't need the acceleration either because if we have a constant speed, remember the acceleration is zero. So we have 80 kilometers per hour, we're gonna convert this to our standard unit of meters per second. So we multiply by 1000 m divided by one kilometer to remove the unit of kilometers. Then we multiply by one hour per 3600 seconds to get rid of that unit of hours. We're gonna be left with meters per second. When we work it out, we actually gonna be 22.22 repeated meters per second. So that's the speed we're going during this reaction phase. And we know that our reaction time is 0.9 seconds. So we spend 0.9 seconds continuing at this constant speed before we actually apply the brakes. And that's the amount of time it takes for us to react. And we wanna find delta Xr now because we have constant speed. No acceleration. Recall that the equation is gonna be that the velocity V is equal to delta X divided by T. What we want to calculate is delta X. And so in this case, delta Xr is going to be equal to VR multiplied by tr substituting in our values, we get that delta Xr is gonna be 22.22 repeated meters per second multiplied by 0.9 seconds. And if we multiply the two together, we get that the amount of time it takes for us to react or sorry, the distance covered during our reaction is 20 m. Ok? So that is delta Xr and we are gonna need that in our equation for the stopping distance. That is the first part. All right. So we traveled 20 m before we actually react to our breaking. All right, now, we're gonna move to the second part and that is when we are decelerating. So this is going to be our deceleration phase and we know that we are gonna start with an initial speed of 80 kilometers per hour and that initial speed is gonna be that constant speed we were traveling before we applied the brakes. So 80 kilometers per hour, which we now know is 22.22 repeated meters per second. We know that the final velocity is going to be zero because we're looking for the stopping distance, the stopping distance mean we means we come to a stop. So our final velocity is zero. We're told that we have a deceleration of negative 2 m per second squared. OK? So acceleration negative 2 m per second squared and we wanna find Delta XD, which is the, the amount or the distance traveled during the deceleration phase. So we're gonna use our U AM or kinematic equations. We have three known values VNVF and a one thing we're trying to find delta X. So we're gonna choose the kinematic equation with those four variables. We can ignore the time T because that's not what we're looking for. We don't have information about it. That equation is gonna be VF squared is equal to V not squared plus two multiplied by a multiplied by delta X substituting in our values zero is equal to 22.22 repeated meters per second, all squared plus two multiplied by negative 2 m per second squared, multiplied by delta XD. We're gonna move the right hand turn to the left hand side and we have negative 4 m per second squared multiplied by DELTA XD when we simplify. So we're gonna add it to both sides. We get 4 m per second squared, multiplied by delta XD is equal to 22.22 repeated meters per second squared. And when we divide by 4 m per second squared, we get that delta XD is gonna be 123.45679 m approximately. OK. So the distance covered while we are actually decelerating is gonna be 100 and 23.45679 m. OK. All right. So we've figured out these two parts. Now, let's remind ourselves of what we're looking for. We're looking for the stopping distance. Delta X and we said that the stopping distance is gonna be made up of the distance we travel during the reaction phase. Ok. So while we react to having to stop, plus the distance it takes to actually stop. Once we've applied the brakes, we found these two values Delta Xr and Delta XD. Now we just need to add them together. So our total stopping distance Delta X is going to be equal to 20 m. What? 123.45 6 7 9 m. And when we work this out, we get that delta X is going to be equal to approximately 143 4 5 6 7 9 m. And that is the total stopping distance comparing this to our answer choices. We are gonna round to two significant digits and write this in scientific notation and we get that the correct answer is option B 1.4 multiplied by 10 to the exponent 2 m. Thanks everyone for watching. I hope this video helped see you in the next one.

Struggling with Physics?

Determine the stopping distances for an automobile going a constant initial speed of 95 km/h in the +𝓍 direction, and human reaction time of 0.40 s:
(a) for an acceleration a = ― 2.5 m/s²"

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Struggling with Physics?

Determine the stopping distances for an automobile going a constant initial speed of 95 km/h in the +𝓍 direction, and human reaction time of 0.40 s:(a) for an acceleration a = ― 2.5 m/s²"

Watch next

Kinematics Equations practice set

Determine the stopping distances for an automobile going a constant initial speed of 95 km/h in the +𝓍 direction, and human reaction time of 0.40 s:
(a) for an acceleration a = ― 2.5 m/s²"