While driving along the highway at 40 m/s, you spot a police car 50 m ahead, traveling at a constant speed of 30 m/s which is the speed limit. You apply the brakes and begin decelerating at 1.0 m/s^2. Assume that the police officer will give you a speeding ticket only if you pass her car. Will you get a ticket?
As with most kinematics problems, there are many different ways to get the right answer. Firstly, it is always helpful to identify what information we have, and what we don't. We have: initial velocity (vi) = 40 m/s final velocity (vf) = 30 m/s initial position (xi) = 0 m acceleration (a) = - 1.0 m/s^2 We don't have time (t) or the final position (xf) where you get to 30 m/s, and it's worthwhile to note that this is one-dimensional. Let's find out how long it's going to take to decelerate to 30 m/s first, then we can compare the distance travelled by both you and the police officer. To do this, we can use our handy kinematics equations, choosing the one with time as the only unknown variable. vf = vi + a (t) 30 m/s = 40 m/s - 1 m/s^2 (t) 30 - 40 = - t -10 = -t t = 10 seconds So, it takes 10 seconds to decelerate to the speed limit. How far do both you and the police officer travel in that 10 seconds? To answer this, we can use another kinematics equation: xf = xi + vi (t) + 0.5 (a) (t^2) xf = 0 m + 40 m/s (10 s) + 0.5 (-1 m/s^2) (10 s)^2 xf = 0 + 400 - 0.5 (100) xf = 400 - 50 xf = 350 m So, it takes 350 meters for you to decelerate. Now, we must find out how far the police officer travelled. Since we only identified the information we had about you, let's take a moment to identify what information we have about the officer, then follow the same steps to find the distance travelled. vi = 30 m/s vf = 30 m/s xi = 50 m xf = ? a = 0 m/s^2 t = 10 s (Since we must not pass the officer in those 10 seconds where we are above the speed limit) xf = xi + vi (t) + 0.5 (a) (t^2) xf = 50 m + 30 m/s (10 s) + 0.5 (0 m/s^2) (10 s)^2 xf = 50 + 300 + 0 xf = 350 m So, the police officer also travels 350 meters in those 10 seconds. Therefore, you catch up to the officer, but you don't pass her! Ticket averted! That was a close one. Better stay closer to that speed limit next time!
Find the vertex of the parabola. y = 2 x^2 - 8x + 7
To find the vertex of any parabola (aka quadratic function), we can take advantage of vertex form: y - k = a (x - h)^2, where a, h, and k are constants and the coordinates (h, k) are the vertex of the parabola. If we can manipulate the function and put it into vertex form, we can identify the vertex by finding h and k. Let's focus on getting the right hand side of the equation first. How can we turn 2 x^2 - 8x + 7 into the form a (x - h)^2? The technique called completing the square will help us do just this. First, let's send that 7 over to the other side: y - 7 = 2 x^2 - 8x Next, we should factor out the 2 in front of the x^2, this will become our constant a. y - 7 = 2 (x^2 - 4x) Then use whatever constant is in front of the x-term (in this case -4), divide it by 2, square it, then both add and subtract it after that term (this keeps our equation the same so that the net change is 0). -4 --> -4 / 2 = -2 --> (-2)^2 = 4 y - 7 = 2 (x^2 - 4x + 4 - 4) Now, notice that x^2 - 4x + 4 factors into (x - 2)^2. The only term in our way from doing that is the -4, but to send it to the other side we must be mindful that it is still inside the parenthesis to be multiplied by 2 (variable a). Therefore, we have to take it out of the parenthesis by multiplying it by the 2 and then adding it to the other side. y - 7 = 2 (x^2 - 4x + 4 - 4) y - 7 = 2 (x^2 - 4x + 4) + 2 (-4) y - 7 = 2 (x^2 - 4x + 4) - 8 y - 7 + 8 = 2 (x^2 - 4x + 4) y + 1 = 2 (x^2 - 4x + 4) Finally, factor the right hand side to finish putting our equation into vertex form. y + 1 = 2 (x - 2)^2 Being mindful of the signs in front of the h and k in vertex form, we can see that h = 2, k = -1, and a = 2. Therefore, the vertex of the parabola is at (2, -1)!
Find the limit if it exists. lim x->0 [2 x tan(x) / sin(x)]
The first step when finding limits is to plug the limit into the variable. Therefore, in this case we must plug 0 into x: What happens to each term? 2x -> 0 tan(x) = sin(x) / cos(x) -> 0 / 1 sin(x) -> 0 So, we get 0 / 0, an indeterminate form. What do we do now? L'Hospital's Rule tells us that if we get an indeterminate form when doing a limit, take the derivative of the numerator and the derivative of the denominator and try the limit again. So, to take the derivative of the numerator we must use the product rule because there are two instances of the variable (x) being multiplied by each other: d/dx (2 x tan(x)) = (2 x) (sec^2(x)) + (2) (tan(x)) Then the denominator: d/dx (sin(x)) = cos(x) Now, take the limit again looking at each term: lim x->0 [(2 x sec^2(x) + 2 tan(x)) / cos(x)] 2x -> 0 sec^2(x) = 1 / cos^2(x) -> 1 2 tan(x) -> 2 * (0 / 1) = 0 cos(x) -> 1 So, we get lim x->0 = (0 * 1 + 0) / 1 = 0 / 1 = 0 The limit is zero!