What Are the Speeds of Two Cars Meeting on a Highway?

  • Thread starter Thread starter ajay.05
  • Start date Start date
  • Tags Tags
    Motion
AI Thread Summary
Two cars start from points A and B, which are 100 km apart, and meet after 5 hours when traveling in the same direction and after 1 hour when traveling towards each other. The speeds of the cars can be represented as x and y, with x being the faster speed. To solve for their speeds, two equations are established based on their positions over time, considering the distance covered in each scenario. The first equation accounts for the distance traveled when moving in the same direction, while the second equation addresses the distance when they are approaching each other. By setting up these equations, the speeds of the two cars can be determined mathematically.
ajay.05
Messages
46
Reaction score
3

Homework Statement


Places A and B are 100 km apart on a highway. One car starts from A and the other from B, at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in hour. What are the speeds of two cars?

Homework Equations


http://sketchtoy.com/63036238

The Attempt at a Solution


In the first case,
I assumed that both are traveling toward a point C, and the speeds are x,y respectively(x>y).
Therefore, If car 2 were made to be still, then car 1 will travel with a speed of x-y
So, x-y=20 (Speed = Distance/time)

How can I continue with Case 2?
Help me out:)
 
Last edited:
Physics news on Phys.org
Hello Ajay, and welcome to PF.

:smile: I liked the sketchtoy. But what you really need are some relevant equations.

For the first case:
One for the position of car A as a function of time, and one for the position of car B. Position of point C is where the cars are at the same place. Where that is is not relevant. But you know that at time t=5 hours the equality holds.

Second case same story, different expression for one of the two positions, different C (again, irrelevant). But at time t = 1 hour the equality holds.

And now you have two equations with two unknowns, the speeds of the cars. Presto, physics = math !
 
I would keep to the standard approach.

When traveling in the same direction, the distance each covers is its speed x duration. That gives you one equation.

What traveling in opposite directions, the distance each covers is its speed x duration. The other equation.

Two equations, two unknowns.
 
I assumed that both are traveling toward a point C
is a good start. You don't know the speeds, so you name them va and vb. Position in one dimension can be given as x. And a reference point can be xA=0, so that xB=100 km.

So, starting from your relevant equation x(t) = x(0) + v * t
-- this is how you can write your Speed = Distance/time in a convenient way for later use ! --
(this way it has the look of an equation, as opposed to "conversation" :smile:)​
you write down the position of car a as a function of time:
xa(t) = ...​
 
Hi Ajay,

I would start by adding some additional detail to your sketch. Draw out each situation on an X-axis with some labels for your distance, and vector arrows labeled with the velocities va and vb.

Then work on creating the position equations for each car in each situation, as BvU suggested.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top