Distance traveled and displacement Suppose an object moves alo...

Question

Distance traveled and displacement Suppose an object moves along a line with velocity (in ft/s) v(t)=6−2t, for 0≤t≤6, where t is measured in seconds.

c. Find the distance traveled by the object on the interval 0≤t≤6.

Accepted Answer

Welcome back everyone. In this problem, a particle moves along a straight path with a velocity V of T equals 10 minus 5 T in feet per second fort greater than or equal to 0 but less than or equal to 4. What is the total distance the particle travels during this interval? A says is 12 ft, B 10 ft, C 16 ft, and D 20 ft. Now what do we know that will help us to find the total distance given a formula for the velocity? Well, recall that velocity is the integral of distance. Or sorry, distance is the integral of the velocity. So a distanced would be equal to the integral of the absolute value of our velocity function with respect to time, which means it's going to be the definite integral between 0 and 4 of the absolute value of 10 minus 5 T with respect to T. So if we can integrate 10 minus 5T with respect to T, then we should be able to find the distance. But no, let's analyze or integrate. Notice that when 10 minus 5 T is equal to 0, the term inside our absolute value function, then 10 equals 5T, which means T would be equal to 2. Thus this tells us that 10 minus 5 T is greater than or equal to 0 when T is between 0 and 2. And 10 minus 5 t then would be less than 0 when T is greater than 2 but less than or equal to 4. So that means we can or we'll have to split our integra at T equals 2. In other words, the distance is going to be equal to the integral between 0 and 2 of 10 minus 5 T with respect to T. Plus, the integral between 42 and 4, sorry, of 10 minus 5 T. OK, with respect to T, where in this case, when it's between 2 and 4, 10 minus 5 T will be less than 0. If that's the case, that means here we could rewrite this then. As our integral between 2 and 0 of 10 minus 5T with respect to T plus our integral between 2 and 4 of 5 T minus 10 with respect to T because if 10 minus 5T is less than 0, then this is what that would imply. So now let's integrate both of these separately. No, for for integral between 0 and 2 we know that the integral of 10 is 10T. The integral of negative 5T would be -5 house of T squared and now all of this is between 0 and 2. On the other hand, Again, the integral of 5T is 5 halves of T squared. OK. Integral of -10 is -10 T and that would be between 2 and 4. Now for the first integral, OK, we know that if we substitute 0, the entire result will be 0. So if we substitute the upper limit, that's going to be 10 multiplied by 2 minus 5 halves of 2 squad. For the other part of our integral here, OK. Then we're going to have 5 halves of 4 squared, OK, minus 10 multiplied by 4. And Minus 5 halves of 2 square minus 10 multiplied by 2, OK. I know when we go ahead and solve in our first term, 10 multiplied by 2 is 20. 5 halves of 2 squared would be 10. OK. So that's 20 minus 10 plus. 40 minus 40, OK, minus 10 minus 20. No, when we go ahead and simplify this, 20 minus 10 is 1040 40 is 0, and 10 minus 20 is -10, so negative -10 would be positive 10. So really that's 10 + 10. Thus that tells us our distance is 20 ft. The distance traveled is 20 ft. Thanks for watching everyone. When we look back at our answer choices, that means that D is the correct answer.

Distance traveled and displacement Suppose an object moves along a line with velocity (in ft/s) v(t)=6−2t, for 0≤t≤6, where t is measured in seconds.c. Find the distance traveled by the object on the interval 0≤t≤6.

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Distance traveled and displacement Suppose an object moves along a line with velocity (in ft/s) v(t)=6−2t, for 0≤t≤6, where t is measured in seconds.

c. Find the distance traveled by the object on the interval 0≤t≤6.