Chase problem involving acceleration and velocity

AI Thread Summary
Car A accelerates from rest at 0.75 m/s² for one minute, reaching a speed of 45 m/s and covering 1350 m. Car B accelerates from rest at 0.1 m/s² but has not yet caught up with Car A by the time calculations were made. The total distance covered by Car A after 450 seconds is 17900 m, indicating that Car B has not yet reached this distance. The correct approach to solve the problem involves equating the distances of both cars after the initial acceleration phase. Accurate formulas for distance and velocity must be used to find when Car B will catch up.
patton_223
Messages
9
Reaction score
0

Homework Statement



Car A accelerates from rest at 0.75m/s^2 for 1 minute. After this Car A cruises at a constant speed. Car B accelerates from rest at the same time at 0.1m/s^2. When will car B catch up with Car A? What distance have they traveled and what is the final speed of each car?



Homework Equations


d= ViT + 1/2at^2
Vf= Vi + aT

The Attempt at a Solution


i found the speed of car A which is 22.5 m/s and Car B which is 45 m/s

and i did the calculations and i got 450 sec for the time for Car B to catch up with car A

but when i tried to find the distance i got really odd numbers, i got 10125m

because i would plug in 450 sec into (1/2)(0.1)(450^2) and got that

i just want to know if i did this right, and i could use some help if this is wrong
 
Physics news on Phys.org
Welcome to PF, Patton!
How did you find the speed of car A after 60 seconds?
I used V = Vi + a*t = 0 + .75*60 and got 45 m/s.
No velocity for car B can be found because we have no time for it . . so far.

The big picture on a chase problem is to write
car A distance = car B distance
Then you just need a complete formula for each distance and solve the equation for time. I'm pretty sure the meet will be after the initial 60 seconds so you only need formulas valid after that time. If you have a go at the two formulas, we'll be glad to help you make sure they are correct.

I checked your time of 450. At that time car A has gone the initial .5*.75*60² = 1350 m plus the d = v*t = 45*(450-60) = 17550 m for a total of 17900 m. Looks like car B still has a long way to catch up at time 450!
 
Delphi51 said:
Welcome to PF, Patton!
How did you find the speed of car A after 60 seconds?
I used V = Vi + a*t = 0 + .75*60 and got 45 m/s.
No velocity for car B can be found because we have no time for it . . so far.

The big picture on a chase problem is to write
car A distance = car B distance
Then you just need a complete formula for each distance and solve the equation for time. I'm pretty sure the meet will be after the initial 60 seconds so you only need formulas valid after that time. If you have a go at the two formulas, we'll be glad to help you make sure they are correct.

I checked your time of 450. At that time car A has gone the initial .5*.75*60² = 1350 m plus the d = v*t = 45*(450-60) = 17550 m for a total of 17900 m. Looks like car B still has a long way to catch up at time 450!

I found the speed of car a by using d=1/2aT^2 i used the acceleration and 60sec and found the distance to be 1350m and then divided that by 60sec to find the velocity of 22.5

but it's the distance that's throwing me off
 
I found the speed of car a by using d=1/2aT^2 i used the acceleration and 60sec and found the distance to be 1350m and then divided that by 60sec to find the velocity of 22.5
That 1350 m for the first 60 seconds is correct. But you used d = vt to find the velocity and d=vt does not apply to this portion of the motion because it is accelerated. You must use V = Vi + at to find the velocity at time 60.
 
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}...
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...
Back
Top