At t=0t=0, a car approaching a stop sign decelerates from a speed of 50 mi\operatorname{mi}/hrhr according to the acceleration function a(t)=4t+3a\left(t\right)=4t+3, where t≥0t\ge0 and is measured in hours. How far does the car travel between t=0t=0 and t=0.1t=0.1 hrhr?

5.02 mi ⁡ 5.02\operatorname{mi}

At t = 0 t=0 , a car approaching a stop sign decelerates from a speed of 50 mi \operatorname{mi} / h r hr according to the acceleration function a ( t ) = 4 t + 3 a\left(t\right)=4t+3 , where t ≥ 0 t\ge0 and is measured in hours. How far does the car travel between t = 0 t=0 and t = 0.1 t=0.1 h r hr ?

So in this problem, we are told at time t equals zero, a car approaching a stop sign decelerates from a speed of 50 miles per hour. Now we're also given the acceleration function a of t, where t is the time greater than or equal to zero and is measured in hours. We're asked how far does the car travel between this time interval right here, zero to zero point one hours. So this is an interesting problem with a lot of details here, and where we should start is by figuring out what exactly we have in this problem and where we're trying to go. Now something I notice right off the bat is this acceleration function really sticks out to me. Because if we're given a function for some type of motion, in this case acceleration, that tells me that that's gonna be a good starting place. So we have four t plus three. Now something I also note is that the start of the problem here, we have that time t equals zero gives us a speed of 50 miles per hour. What does that mean? Well, what that means is that our initial speed, v of zero, is going to equal 50 miles per hour, because if at time t equals zero, this is the speed, then that's our initial speed. So we have an initial speed, and we have this acceleration function. Now what we're trying to calculate is how far the car travels between the time of zero and the time of zero point one hours. So if I want to calculate this, well, that sounds like I'm looking for a displacement if we're figuring out how far this car travels. But I can't just find the displacement directly from acceleration. So what I first need to do here is figure out what my velocity is. So I can integrate my acceleration to get velocity, then I can integrate the velocity to get position, or change in position to figure out how far the car has traveled. So that's gonna be my strategy for solving this. So my first step is gonna be figuring out the velocity, v of t. And recall we have an equation that allows us to do this. V of t is gonna be equal to the initial velocity, plus the integral from our initial time to some time t of our acceleration function as a function of x. We need to use a dummy variable here. So if I wanna find my velocity as a function of time, what I can do is I can add my initial velocity, which is v of zero, to the integral from my initial time of zero to some time t, this variable here, of my acceleration function as a function of x. So that's just gonna be taking this t here and replacing it with x and then integrating with respect to x. So now we have our integral and our problem set up in order to find our velocity. Now the initial velocity is given to us as 50. And then to integrate this, well, I can just use the rules for calculus that I already know about. So you take the antiderivative of four x, which using the power rule will give me four x squared over two. Next, I can take the antiderivative of three, which is 3x, and all of this is going to be bounded from zero to t. Now at this point, notice we have a zero as our lower bound. If we plug zero in for x, it's just going to clear everything inside these brackets. So we don't need to worry about this lower bound. We only need to worry about the upper bound t. So what I can do is first reduce any fractions I see. So I have this 50 plus and then four divided by two. That's going to give us two x squared. Then we have plus three x, and then this is all bounded from zero to t. And then we can go ahead and plug in this t directly. So plug in this t into here, and that's going to give me two t squared. So we've gone ahead and dealt with this term. Then next, I'm gonna take our t and plug it in for the x that's being multiplied by three, and that's gonna give me three t. So now we've dealt with that term. Now lastly, I'll just go ahead and latch on this 50 here, so we have plus 50, and this will deal with the last term. So now we've gone ahead and found our velocity function, v of t, and that is the function we're looking for for this first part of the problem. Now I'm not quite finished yet because our goal was not to find the velocity function. Our goal was to figure out how far the car had traveled in this time here, which means we're actually looking for a displacement. But recall that when calculating displacement, we actually do need the velocity function, because the displacement is gonna be the integral of our velocity function with respect to time based on the time interval t initial to t final. So if I want to calculate my displacement for this specific problem, well, what I'm gonna do is integrate my velocity function. Now the velocity function is given right here as two t squared plus three t plus 50, and we're integrating that from zero all the way up to 0.1. And this will allow us to find our displacement. So let's go ahead and do this integral. Now what I can do is use the rules for calculus that I already know about. I can take the antiderivative of 2t squared, which using the power rule, be 2t cubed over three. Then we're going to have plus the integral of three t, which is gonna be three t squared over two, plus then we'll integrate 50, which will give us 50 t. And then we're gonna go ahead and bound this whole thing from zero to 0.1. Now notice that this zero here would just set everything in these brackets equal to zero, because everything is multiplied by a t here. So I only need to focus on the 0.1 in the top bound. So doing this, we're going to have two times 0.1 cubed over three, plus three times 0.1 squared over two plus 50 times 0.1. Now you can go ahead and check these numbers on a calculator here, and you should get a value of 5.02. So five point o two, and then that's gonna be in miles, which would be the solution to this problem. The reason I know this is in miles is because our problem gives us miles per hour that we're dealing with. Our time is in hours, and then our distance is in miles. So I can tell my displacement would be five point o two miles. So this here would be the solution, and that is how you can solve this problem. So I hope you found this video helpful. Let me know if you have any questions, and let's move on.

At t = 0 t=0 , a car approaching a stop sign decelerates from a speed of 50 mi ⁡ \operatorname{mi} / h r hr according to the acceleration function a ( t ) = 4 t + 3 a\left(t\right)=4t+3 , where t ≥ 0 t\ge0 and is measured in hours. How far does the car travel between t = 0 t=0 and t = 0.1 t=0.1 h r hr ?

5.02 mi ⁡ 5.02\operatorname{mi}

0.02 mi ⁡ 0.02\operatorname{mi}

50.32 mi ⁡ 50.32\operatorname{mi}

0.32 mi ⁡ 0.32\operatorname{mi}

10. Physics Applications of Integrals

Kinematics

Calculus

10. Physics Applications of Integrals

Kinematics

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Multiple Choice

At $t equals 0$ , a car approaching a stop sign decelerates from a speed of 50 $mi$ / $h r$ according to the acceleration function $a (t) = 4 t + 3$ , where $t is greater than or equal to 0$ and is measured in hours. How far does the car travel between $t equals 0$ and $t equals 0.1$ $h r$ ?

$0.02 mi$

$50.32 mi$

$5.02 mi$

$0.32 mi$

Verified step by step guidance

Step 1: Recognize that the problem involves finding the distance traveled by the car, which requires integrating the velocity function over the given time interval [0, 0.1]. The velocity function v(t) can be obtained by integrating the acceleration function a(t).

Step 2: Write the acceleration function a(t) = 4t + 3. To find the velocity function v(t), integrate a(t) with respect to t. This gives v(t) = ∫(4t + 3) dt = 2t² + 3t + C, where C is the constant of integration.

Step 3: Determine the constant of integration C by using the initial condition. At t = 0, the car's velocity is 50 mi/hr. Substitute t = 0 and v(0) = 50 into the velocity equation: 50 = 2(0)² + 3(0) + C. Solve for C to find C = 50.

Step 4: Substitute C back into the velocity function to get v(t) = 2t² + 3t + 50. Now, to find the distance traveled, integrate the velocity function over the interval [0, 0.1]. The distance traveled is given by s = ∫[0, 0.1] v(t) dt = ∫[0, 0.1] (2t² + 3t + 50) dt.

Step 5: Perform the integration step-by-step. Break the integral into parts: ∫[0, 0.1] (2t²) dt + ∫[0, 0.1] (3t) dt + ∫[0, 0.1] (50) dt. Compute each integral separately, evaluate the definite integrals, and sum the results to find the total distance traveled.

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