An antelope moving with constant acceleration covers the distance...

Question

An antelope moving with constant acceleration covers the distance between two points 70.0 m apart in 6.00 s. Its speed as it passes the second point is 15.0 m/s. What is (b) its acceleration?

Accepted Answer

welcome back everybody. We have a cheetah that is chasing a deer and I'm going to picture this cheetah is just a little head with ears. Now as he is chasing this deer, he passes through the entire distance between two poles. Now the distance between these two poles we are told Is 100 m and we are told that he does this in a matter of 10 seconds. We are also told that the speed at which he passes the second pole is 15 m/s and we are tasked with finding what his velocity was when he passed the first pole. Now I see some velocities, I see time and I see a change in position that leads me to using the Kinnah Matic formula. So let's see if this one works. We have a kid a magic formula that says our final velocity equal to our initial velocity plus our acceleration times time. But we don't know our initial velocity, but we also don't know our acceleration. So this equation has two unknowns, We can't use it by itself. So what about if we use our final velocity squared is equal to our initial velocity squared plus two times our acceleration times delta S. Well, once again we have an equation with two unknowns. But since we have a cyst of two equations with two unknowns, we can use both of these systems to figure out our initial velocity. So let's go ahead and do that first and foremost, I'm going to look at this first equation, which I'm gonna deem equation one and plug in what we know. Well we know our velocity when he passes the second pole or final velocity is 15. So I'm gonna say 15 is equal to our initial velocity, you don't know plus our acceleration times our time which is just 10 seconds. Now, what I'm gonna do just so that these variables are on separate sides, I'm going to subtract V O from both sides. This gives us that 10 times. Our acceleration is equal to 15 minus V. O. Can't really do anything else from here. So let's address our second equation. So our final velocity squared is 15 squared, which is equal to our initial velocity squared, which is trying to find plus two times are acceleration times 100. And when we simplify some things, we get that 225 equal to our velocity squared plus this term right here and I'm gonna do something a little tricky here. This term is equal to 200 times our acceleration. But if I take out a factor of 20, we are left with 20 times 10 times our acceleration. But we know what 10 times our acceleration is. We know that it's 15 minus V. O. So I'm going to plug this into here and here is what we get from that. We get that our initial velocity squared plus 20 times 15 minus our initial velocity is equal to 25. I'm gonna go ahead and distribute this 20 here, which gives us that V O squared plus minus times. Our initial velocity is equal to 2 25. Subtract 2 25 Both sides. Which yields that our initial velocity squared plus 75 minus 20. V zero is equal to zero. I'm actually gonna move these terms around a little bit to get it into its quadratic form. I'm gonna say that V O squared minus 20. V O plus 75 is equal to zero. And now we can factor this equation, we get that zero minus times. Zero minus five is equal to zero, yielding that are factors for V zero is equal to 15 and five but which one is it? Which one of these values makes sense in this context? Well, You can kind of infer from the problem that our velocity is increasing but our final velocity is already 15. Can't be this. So we are going to say that our initial velocity is five, which is equal to our answer choice of a thank you guys so much for watching. Hope this video helped and we will see you all in the next one

An antelope moving with constant acceleration covers the distance between two points 70.0 m apart in 6.00 s. Its speed as it passes the second point is 15.0 m/s. What is (b) its acceleration?

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