A car is heading along a flat slippery road at a speed of 95 k...

Question

A car is heading along a flat slippery road at a speed of 95 km/h. The minimum distance within which it can stop (because of ABS) is 66 m. Estimate the radius of the sharpest curve the car can negotiate on an icy flat surface at the same speed without skidding out.

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. The enhanced braking system of a bus allows it to stop in a minimum distance of 67 m on a slick and level road while traveling at a speed of 96 kilometers per hour on a frost covered surface. What is the smallest radius value of the sharpest curve that the bus can navigate without sliding off of the road? So that's our end goal is we're trying to figure out the smallest radio, smallest radius value of the sharpest curve that this particular bus can drive on without sliding off the road. And that will be our final answer we're ultimately trying to solve for is what is this radius value? We're also given some multiple choice answers. They're all in the same units of meters. Let's read them off to see what our final answer might be. A is 1.3 multiplied by 10 to the power of two B is 2.2 multiplied by the 10 to the power of two C is 3.3 multiplied by 10 to the power of two D is 4.1 multiplied by 10 to the power of two. Awesome. So first off, let us draw a diagram to better visualize all the forces at work in this particular problem. So I've gone ahead and put together a fancy diagram for us to refer to, to help us visualize this prompt. We have our road represented in black and that represents our frost covered surface and then we have our bus in blue. Uh then it's coming up about the turn. And as you could see, pointing to the left, we have our centrifugal force and then pointing up from the bus, we have our normal force and then pointing down from our bus, which isn't, isn't specified, but that's our weight force. So that's the mass of the bus multiplied by the acceleration due to gravity. So it's M multiplied by G and that's our weight force that's pointing in the downwards direction. And then pointing to the right, we have our frictional force, which is the friction force which is opposing the movement of the bus. So since the bus is going, the friction force is gonna be opposing it. OK. So with that in mind now that we have a better visualization of what's going on in this particular problem, we now need to note that the, the acceleration. So for this acceleration that the static friction can provide. So the static friction that is provided can be determined from the minimum stopping distance. When we assume that the bus is just on the verge of sliding. Let us also recall that for an unbanked curve, the same static frictional force is used to provide the centripetal acceleration. Since the condition of the acceleration of the bus to its stopping point distance is negative, the centripetal acceleration will be in the opposite direction. So now at this point, we need to recall and use the equation to help us determine for the speed of an object which is, and we can call this equation one which states that VF squared is equal to V squared plus two multiplied by a multiplied by parenthesis. XF minus X I parenthesis where VF is our final speed of the bus V I is the initial speed of the bus. A is equal to the acceleration and XF minus X I is the minimum distance that the bus can stop without sliding where XF is the final distance minus X I, which is the initial distance to be specific. So now at this point, we need to rearrange equation one to isolate and solve for a like. So, so when we start the rearranging process, we will get that VF squared minus V I squared is equal to two multiplied by a multiplied by XF minus X I. So getting a all by itself and let's call this equation two, we'll call it A S which we, since we're solving for acceleration, but specifically, this is the stopping acceleration. Let's call this A S to denote the stopping acceleration is equal to VF squared minus V squared all divided by two multiplied by parenthesis XF minus X I Awesome. Where once again A S is the acceleration for the stopping distance. And let us also note that the final speed of the bus will be zero since it will come to a stop, meaning it's not gonna be moving. Therefore, we could rewrite equation two as and let's call this equation three. That A S is equal to negative V squared divided by two multiplied by parenthesis XF minus X I Awesome. So let us note that the acceleration from the stopping distance is negative and is in the opposite direction of the centripetal acceleration. Thus, we can write the following. We can write and let's call this equation for that. A S is equal to negative a subscript capital R where a capital R or I'm just gonna say R to keep it simple. So A R is equal to the sin trippet acceleration. So now at this point, we can substitute this value into equation three and we will get, and let's call this equation five and we'll draw a little line to separate it because this is separate from the equation four. That A R is equal to V squared divided by two multiplied by XF which is parentheses X FX F minus X I parenthesis. OK. Moving right along here. So now we need to recall and write the equation for the centripetal acceleration which we call this equation six. And let us state that A is equal to V I divided by R where R is the radius of the curved path. So now at this point, we need to set equations five and six equal to each other. And let's call this equation seven. So we will take V I squared which I forgot to put a square on our V I in our six, our equation six. So note that A R is equal to V I squared divided by R. That's very important. We need to have the squared value divided by R is equal to V I squared divided by two multiplied by parenthesis. XF minus X I parenthesis. So now at this point, we need to rearrange equation seven to isolate and solve for R. And let's call this equation eight. So equation eight states that R when we use a little bit of algebra to get R by itself, we will get that R is equal to two multiplied by XF which parentheses XF minus X I Awesome. So now at this stage, we can plug in all of our known variables to solve for R, which is our final answer. Let us also quickly note that the minimum distance that the bus can stop without sliding is 67 m. And we can substitute this value in for XF minus X I to help us find the shortest path that the bus can travel without sliding. Therefore, I'll draw a little arrow. Therefore, we can write that R is equal to two multiplied by 67 m, which is equal to 134 m, which is approximately equal to 1.3 multiplied by 10 to the power of 2 m. When we write our final answer in scientific notation, hooray, we did it. So therefore the shortest path that the bus can travel without sliding at the same speed is approximately 1.3 multiplied by 10 to the power of 2 m. And this corresponds with multiple choice answer letter A 1.3 multiplied by 10 to the power of 2 m. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

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A car is heading along a flat slippery road at a speed of 95 km/h. The minimum distance within which it can stop (because of ABS) is 66 m. Estimate the radius of the sharpest curve the car can negotiate on an icy flat surface at the same speed without skidding out.

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