Question

Brakes are applied on a car moving at a speed of 80km/h, and stops in 40m. ¿How many meters had it moved in the instant its velocity fell to:
a) 50km/h
b) 25km/h
c) ¿How long did it take to stop?

Answer 1

Recall the following kinematics equation. (If this is not familiar to you, let me know and I can derive it for you.)\[v_f^2=v_i^2+2a\Delta x\]First use it to find the overall acceleration of the car. Take note that by it going to rest, the final velocity becomes zero. Now, you can use this acceleration to find the displacement for any final velocity by rearranging the equation in the following manner.\[\Delta x = \frac{v_f^2-v_i^2}{2a}\]You can also use the found acceleration value to calculate the time it took for the car to stop by rearranging another kinematic equation.\[v_f = v_i + at \Rightarrow t=\frac{v_f-v_i}{a}\]Let me know if you still need more help.

Answer 2

The equation is not familiar to me, but I can understand the rest. This is actually calculus, don't know if that makes a difference.

Answer 3

No it isn't calculus, it is algebra. You could use calculus but it isn't needed here.

Answer 4

You can derive the first equation in the following manner. Consider the two following kinematic equations of constant acceleration, which I trust you are familiar with. If you do have some background in calculus, you may notice that the second is just the derivative of the first (or that the first is the integral of the second and that the second follows from the definition of acceleration), but that is not important. These you MUST know. Here, a subscript i indicates initial, and a subscript f indicates final.\[x_f=x_i+v_it+\frac{1}{2}at^2\]\[v_f=v_i+at\]Anyway, subtracting initial position from both sides of the first equation gives us the following.\[x_f-x_i=v_it + \frac{1}{2}at^2\]\[\Delta x = v_it + \frac{1}{2}at^2\]\[\frac{a}{2}t^2+v_it-\Delta x = 0\]Solving for time gives the following.\[t=\frac{-v_i\pm\sqrt{v_i^2+2a\Delta x}}{a}\]Substituting this value into the second of the pair of kinematic equations gives us the following.\[v_f=v_i+a\left(\frac{-v_i\pm\sqrt{v_i^2+2a\Delta x}}{a}\right)\]\[v_f=\pm\sqrt{v_i^2+2a\Delta x}\]\[\boxed{v_f^2=v_i^2+2a\Delta x}\ \checkmark\]I use this equation so commonly that I don't bother deriving it each time -- I just memorize it. Usually I don't suggest rote memorization, but this is one of the situations where it'll actually save you a significant amount of time. However, it is still good to know how to derive it in the case that you really find yourself suck with recalling the formula. I hope this was of help!

Answer 5

The significance of this new kinematic equation is that it since it eliminates the variable t, you can find initial and final velocities given only an acceleration and a displacement. Once you get to conservation of energy in your Physics class, you'll look at the following variation of the equation. Here, for these purposes, we will change a to g, the gravitational acceleration.\[v_f^2-v_i^2=2g\Delta x\]Multiplying both sides by mass and dividing by two gives the following.\[\boxed{\displaystyle \underbrace{\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2}_{\text{Change in KE}}=\underbrace{mg\Delta x}_{\text{Chage in PE}}}\]Here, KE indicates kinetic energy and PE indicates potential energy. This is just a preview!