Velocity of a car in order to meet another car coming towards it

In summary, car A will meet car B at 44.5m if it is going at 20km/h and at 76.6m if it is going at 40km/h.
  • #1
PhizKid
477
1

Homework Statement


Car A and Car B and traveling towards each other from 220m apart. If car A travels at 20km/h, both cars will meet at 44.5m. If car A travels at 40km/h, both cars will meet at 76.6m. What is the initial velocity and constant acceleration of car B?

Homework Equations


1D Kinematics

The Attempt at a Solution


For car B, the formulas just involve distance and velocity since we don't have any other information, V^2 = (V_0)^2 + 2a(x - x_0).

Since acceleration is constant, it just becomes V^2 = (V_0)^2, which is meaningless because they will be the same value since acceleration is constant.

If we try some other formula like x = x_0 + (v_0)t + 1/2at^2, we are left with 220 = 175.5t + (v_0)t, which again is meaningless because we don't have time. (Same with every other equation, it is impossible to solve for time also given the information.)

So the problem doesn't seem to have enough information, yet apparently there is a solution...how do you do this?
 
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  • #2
It is unclear what the distance to the meeting point is relative to. The original midpoint?

Apart from this, just treat the initial velocity and acceleration of B as unknowns. The two conditions given translate into two equations with those unknowns. So the problem is fully determined.
 
  • #3
So for the first condition of car A going at 20km/h, I got:

0.0445 - 220 = V_0 * 0.002225 + 1/2a(0.002225)^2 for Car B

For Car A's 40km/h, I got:

0.0766 - 220 = V_0 * 0.001915 + 1/2a(0.001915)^2 for Car B

I tried solving for V_0 with the first equation and got:

(a(0.00000247531) - 219.9555) / 0.002225 = V_0

I plugged that into the second equation and got -119,670,000 km/hr. Which is obviously wrong. What did I do incorrectly?
 
  • #4
Again, it is not clear how the distance to the meeting point is measured. Can you clarify?

Secondly, convert all units to meters and seconds.
 
  • #5
When car A is going at 20km/h, both cars meet at 44.5m, when they are a distance 220m apart.

When car A is going at 40km/h, both cars meet at 76.6m, when they are a distance 220m apart.

The meeting point are just two givens, not measured
 
  • #6
I converted everything to km and h first because m/s don't make sense to me. I can always convert to meters and seconds afterwards

I re-did the algebra and came up with 31,011 km/h, which again seems very unreasonable. Not sure what else to do here
 
  • #7
PhizKid said:
both cars meet at 44.5m
From what?

I converted everything to km and h first because m/s don't make sense to me.
They don't have to make sense to you. They just need to be consistent. Your equations are not.

Regardless, before you understand what those distances mean, you won't be able to solve this.
 
  • #8
voko said:
From what?
From the endpoints of 0m and 220m

voko said:
They don't have to make sense to you. They just need to be consistent. Your equations are not.

Regardless, before you understand what those distances mean, you won't be able to solve this.

I just use it as a reference to check if my solution is reasonable. What should I be understanding?
 
  • #9
PhizKid said:
From the endpoints of 0m and 220m

It does not make sense. When the cars meet, they cannot be at 44.5 m from both the 0 m and 220 m endpoints.
 
  • #10
The spatial equation of car A in the first case:
t = Sa/Va
t = 44.5/(200/36)

The spatial equation of car B in the first case:
Sb = - V.t - at^2/2
Being Sb = 220 - 44.5 and T is the value found above in the spatial equation of car A.
Acceleration and Speed are negative because the car is moving contrary to car A.

The spatial equation of car A in the second case:
t = 76.6/(400/36)

The spatial equation of car B in the second case:
220 - 76.6 = - Vt - at^2/2

Putting the numerical value into the two equations of car B, and approximating the values, we get:
2v + 7a = -41
v + 4a = -22

Solving it, we find:
V = - 10m/s
A = - 3m/s^2

So, the absolute value of both are these: 10m/s and 3m/s^2 (approximately, of course).

But then again, your exercise text is wrong. It must state clear that they meet at 45.5 from car A's start point (that's what I've assumed). Other else, the solution is totally different.
 

1. What is the formula for calculating the velocity of a car?

The formula for calculating velocity is v = d/t, where v stands for velocity, d stands for distance, and t stands for time.

2. How is the velocity of a car affected by its mass?

The mass of a car does not directly affect its velocity. However, a heavier car may require more energy to accelerate to a certain velocity, resulting in a slower acceleration compared to a lighter car.

3. How does the velocity of a car affect its braking distance?

The velocity of a car directly affects its braking distance. The higher the velocity, the longer it will take for the car to come to a complete stop. This is because the car has more momentum, and it takes more force to slow it down.

4. What is the difference between average velocity and instantaneous velocity?

Average velocity is the total displacement of a car divided by the total time taken, whereas instantaneous velocity is the velocity of a car at a specific moment in time. Average velocity takes into account the entire journey, while instantaneous velocity only focuses on a specific point.

5. How does the velocity of a car determine the outcome of a collision with another car?

The velocity of a car plays a crucial role in determining the outcome of a collision with another car. The higher the velocity, the more force and energy will be involved in the collision, potentially resulting in more severe damage and injuries.

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