A particle's velocity is described by the function vₓ =kt² m/s, w...

Question

A particle's velocity is described by the function vₓ =kt² m/s, where k is a constant and t is in s. The particle's position at t₀ = 0 s is x₀ = ─9.0 m. At t₁ = 3.0 s, the particle is at x₁ = 9.0 m. Determine the value of the constant k. Be sure to include the proper units.

Accepted Answer

Hey, everyone. Welcome back in this problem. The velocity of a ball left in the wind is described by the expression V X is equal to three, multiplied by C multiplied by T squared in meters per second time is measured in seconds. And C is a constant at T knott equals zero seconds. The ball has an initial position X knot equals negative 8.2 m at a later time. T one equals 4.1 seconds. The ball is located at X one equals 8.2 m. We're asked, what is the value of the constant expressed with the right units? OK. So we're given a velocity here. We're also given information about position and time. And so your first instinct might be, let's tackle this with kinematics. Let's use our kinematic equations. However, we have to be careful. OK. We're given VX the velocity equal to three C T squared, which means that the acceleration A X can recall that the acceleration is the derivative of the velocity. And if we take the derivative of B X and the exponent of two comes down and multiplies to our constant of three, and we get six multiplied by C multiplied by T. And what you'll notice at this is that this is not constant and if it's not constant, we can't use our U AM or kinematic equations. OK. Uniform, uniformly accelerated motion does not apply. So what do we do instead? Well, just like velocity is related to acceleration through the derivative, it's also related to the position through an inter So we can write the change in position is equal two. Can I recall the integral from Tino to T one of our velocity V X of T D T substituting in the values we have in our problem. And we're gonna have the initial time T which is equal to zero seconds up to the final time, 4.1 seconds of our velocity three multiplied by C multiplied by T squared D T. And so we have delta X is equal to when we take the integral, OK, we're integrating with respect to T, we have a polynomial T squared, we're gonna raise the exponent by one and then divide by that new exponent, we end up with three C T Q divided by three. So we have C multiplied by T Cute. And this is going to be evaluated from T equals zero seconds or lower bound to T equals 4.1 seconds or upper bound. And now we're getting somewhere, we have this change in position and we can find a change in position because we're given information about the position at X X nought. And the position X one, we can calculate T cubed evaluated at these two end points. And the only thing we're gonna have left in this equation is C so we're gonna be able to solve for that constant C. So delta X recall is X one minus X, not the change in position. This is going to equal to c multiplied by 4. cubed minus C multiplied by zero cubed. The position at X one is 8. m. So we have 8.2 m minus X knot is negative 8.2 m. So 8.2 m minus negative 8.2 m is equal to C Multiplied by 4.1 And this is 4. seconds. Cute. A zero Cubed goes to zero. On the left hand side, we can simplify to 16.4 m. On the right hand side, we have c multiplied by 68.921 seconds. Cute. We can divide and we get that C is equal to 0.238, we have meters divided by seconds cubed. So the unit is going to be meters per second cubed and that is gonna be the value of the constant with the units. Now, if we wanted to double check our units, we can do that. Let's go back up to our V. OK. Now we add V X is equal to three multiplied by C multiplied by T squared. And we know that that has a unit of meters per second. I'm gonna go back down here for a minute and do some dimensional dimensional analysis. We have V X is equal to three multiplied by C multiplied by T squared. We know the dimensions of V X are meters divi and bye seconds. And I'm using square brackets here to indicate that we're talking about dimensions. This is equal to three, which doesn't have a dimension and the unit of C, which we're trying to find the unit of time, which is seconds squared. And just like we saw a regular equation, we can divide by the unit of second squared. And this is gonna give us that the unit of C or the dimension of C is meters divided by seconds cubed, which is exactly what we found when we worked out the solution. OK. So that's just a nice little check to make sure we didn't make any mistakes. Now, if we compare what we found to the answer choices, we found that the value of the constant is 0. m per second cubed, which corresponds with answer choice. C thanks everyone for watching. I hope this video helped see you in the next one.

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A particle's velocity is described by the function vₓ =kt² m/s, where k is a constant and t is in s. The particle's position at t₀ = 0 s is x₀ = ─9.0 m. At t₁ = 3.0 s, the particle is at x₁ = 9.0 m. Determine the value of the constant k. Be sure to include the proper units.

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