When will the 2 cars Have equal speeds? [closed]

Question

If car1 started at 20m, with initial velocity of -4m/s and acceleration of 3 m/s^2. Car 2 started at 15m, with initial velocity of 6m/s and acceleration of 0. At what time will the 2 cars have equal speeds.

Here's my work below:

assuming constant acceleration, we know that $v_{avg}=\frac{\left(v_f+v_i\right)}{2}=\frac{x}{t}$ and the acceleration $a=\frac{v_f-v_i}{t}$ with a little bit of manipulation of the two equations, I got $v_f=\frac{x}{t}+\frac{a}{2}t$ since we need to find the time at which the two cars will have the same speed. That's v_1f = v_2f

then, we get $\left(v_{f_2}=\frac{x_2}{t}+\frac{a_2}{2}t\right)=\left(v_{f_1}=\frac{x_1}{t}+\frac{a1}{2}t\right)$ with manipulation of the above equations for t I get $t=\sqrt{\frac{2\left(x1-x2\right)}{a2-a1}}$

Is this the right approach?

Answer 1

It is much simpler than what you attempted. You don't need to take into account the positions of the cars, since what they ask you is to compute when the velocities of both of them will be equal, and they give you their initial velocities and their accelerations. If you know the initial velocity of a car and its acceleration, you can compute its velocity at any moment in time, and that is independent of whether the car is passing through your street or in another town (its position doesn't matter).

If $v_0$ is the initial velocity of an object, $a$ its constant acceleration, $t$ is the moment in time and $v$ the velocity of the object (which will obviously change with time unless the acceleration is zero), then:

$$v=v_0+a\cdot t$$

Writing an equation like this for each one of your cars, we have:

$$\begin{cases}v_1=v_{0,1}+a_1\cdot t \\ v_2=v_{0,2}+a_2\cdot t\end{cases}$$

And what is the value of $t$ for which the velocities $v_1=v_2$ will be equal? Well, we just have to make $v_1=v_2$ in the equations above:

$$v_1=v_2\quad\Rightarrow\quad v_{0,1}+a_1\cdot t=v_{0,2}+a_2\cdot t\quad\Rightarrow\quad t=\dfrac{v_{0,2}-v_{0,1}}{a_1-a_2}=\dfrac{6-(-4)}{3-0}=\dfrac{10}{3}=3,333$$

At $t=3,333$ seconds, both cars will be rolling at the same velocity (note that this velocity will be 6 m/s, since that is the initial velocity of the second car, which has zero acceleration, and this means that its velocity will remain constant).

Edit: I'll try to clarify here why the positions aren't relevant. To understand this, you just have to think about the physical meaning of the concepts involved: velocity and acceleration.

• Velocity is easy, its units are $m/s$, meters per second, which means that if the velocity of an object is, for example, 3 $m/s$, then that object moves 3 meters in one second.

• In the case of acceleration, you will have noticed that its units are $m/s^2$, meters per second squared. A second squared probably doesn't make much sense in an intuitive way, but what if we write it like this?

$$\dfrac{m}{s^2}=\dfrac{m/s}{s}$$

An acceleration is then expressed in meters per second, per second. This means it measures in how many meters per second a velocity changes in a second.

So, if a car is initially moving with a velocity of 6 $m/s$, and it has a constant acceleration of 2 $m/s^2$, then its velocity will increase in 2 $m/s$ with every second that passes. After one second, it will be 8 $m/s$, after two seconds it will be 10 $m/s$, and so on. Since the problem just asks you when the velocities of the two cars will be equal, and we just reasoned that all you need to know to compute the velocity of a car that moves with constant acceleration is its initial velocity, the value of the acceleration and how much time has passed, we conclude that the positions aren't relevant to calculate the velocities.

Answer 2

The equation for $t$ is worked out correctly, but it doesn't help you get to the solution. The question asks when the cars will have the same speed. Their initial positions do not matter, so they should not show up in your answer.