What is the time taken for these 2 cars to meet?

  • Thread starter Thread starter nyrychvantel
  • Start date Start date
  • Tags Tags
    Cars Time
AI Thread Summary
Two cars, P and Q, start 1400 meters apart, with P traveling at a constant speed of 15 m/s and Q accelerating from rest at 5 m/s². To find the time taken for them to meet, one can set up equations for their positions over time and equate them. Car P's position is given by S = vt, while Car Q's position can be calculated using S = ut + 0.5at², with adjustments for direction. The discussion emphasizes the importance of using kinematic equations rather than graphical methods to solve the problem efficiently. The solution involves calculating the distance each car travels before they meet.
nyrychvantel
Messages
14
Reaction score
0

Homework Statement



Two cars, P and Q, are 1400m apart.
P is traveling at 15 ms-1, Q is accelerating from rest at 5.0 ms-2
They are traveling opposite direction(i.e. they are facing each other), what is the time taken for them to meet?
Also, what is the distance traveled by each car before meeting each other?

Homework Equations



a = \frac{v-u}{t}


The Attempt at a Solution



I got a = 5t, but it did not seem to help much...
Do I have to draw a graph for this question?
 
Physics news on Phys.org
nyrychvantel said:

Homework Statement



Two cars, P and Q, are 1400m apart.
P is traveling at 15 ms-1, Q is accelerating from rest at 5.0 ms-2
They are traveling opposite direction(i.e. they are facing each other), what is the time taken for them to meet?
Also, what is the distance traveled by each car before meeting each other?

Homework Equations



a = \frac{v-u}{t}


The Attempt at a Solution



I got a = 5t, but it did not seem to help much...
Do I have to draw a graph for this question?

No, just write out the position dependence over time for the two cars and equate. Assume Car 1 to b at x=0 and the second car to be at x=1400
Use
S=u*t+0.5*a*t*t (for the accelerated motion) (Use negative sign for acceleration)
S=v*t (for the constant velocity motion)
 
Thank you!
I forgot I can use area under a V-T graph of constant acceleration (and sub v=at) to get the total distance traveled by car Q.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top