A speed skater moving to the left across frictionless ice at 8.0 m/s hits a 5.0-m-wide patch of rough ice. She slows steadily, then continues on at 6.0 m/s. What is her acceleration on the rough ice?

Video transcript

Hey everyone. Welcome back in this problem. We have a box sliding at a constant speed of 5m/s on a frictionless surface enters a rough concrete surface. When the box moves three m on concrete, its speed drops to four m/s were asked to determine the magnitude of deceleration of the box on concrete. The answer choices were given our a one m per second squared. B 1.5 m per second squared, C three m per second squared and D nine m per second squared. Now we have information about some speeds and distances and this is going to be a cinematic type problem, okay or a you am problem uniformly accelerated motion. So let's write out the variables we have and see what we can do to find this answer. Now, we're told that the box is sliding at a constant speed of five m/s On a frictionless surface when it enters the rough concrete surface. So our initial speed v naught is equal to 5m/s. Our final speed. VF Well, let's keep reading The speed drops to 4m/s or final speed is going to be four m/s. We're told that the box moves three m on the concrete. So our delta X value the displacement or the distance that it travels is three m and what we're trying to find is the magnitude of the deceleration. And so what we're trying to find is that acceleration value A and we aren't given any information about the time T so we have three variables that we know V not V F in delta X. So we can use our equations to get our unknown A. This is a simple problem where we just need to plug in these values to the right equation. And the correct equation is going to be the one that doesn't include time T. And so we have the following V F squared is equal to V naught squared Plus two, multiplied by a multiplied by Delta X substituting in our values, we have four m per second squared is equal to five m per second squared plus two multiplied by our acceleration. A multiplied by three m. On the left hand side, four squared, we have 16 and the unit is meters squared per second squared. On the right hand side, we get 25 m squared per second squared Plus 6m multiplied by the acceleration we want to isolate for A. So we can move the 25 m per second meters squared per second squared to the left hand side. We get negative nine m squared per second squared is equal to six m times A and we divide by six m to get an acceleration of negative 1.5. And the unit is meter squared per second squared divided by meters. So we get meters per second squared, which is a unit we expect for acceleration. And so our acceleration is negative 1.5 m/s squared. Now remember that the question was asking for the magnitude of the deceleration. Okay. So the negative in our acceleration indicates that it is a deceleration. The magnitude is just going to be 1.5 m per second squared, the absolute value or the magnitude of that acceleration we found. And so the correct answer here is going to be answer choice B 1.5 m per second squared. That's it for this one. Thanks everyone for watching. See you in the next video.