Same uniform acceleration question.

Andwill117

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Hi this is the same question I can’t get my head around.

reference: https://www.freemathhelp.com/forum/...ion-problem-i-can’t-solve.123133/#post-499779

Two cars are driving at the same constant speed on a straight road, with car 1 in front of car 2. Car 1 suddenly starts to brake with constant acceleration and stops in 10m. At the instant car 1 comes to a stop, car 2 begins to brake with the same acceleration. It comes to a halt just as it reaches the back of car 1. What was the seperation between the cars before they started braking?

All I want to know are enough values to make the same amount equations as there are variables.
Did I miss something or was the question not written properly?

ive attached past attempts.

Thank you
 

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Hi this is the same question I can’t get my head around.

reference: https://www.freemathhelp.com/forum/threads/uniform-acceleration-problem-i-can’t-solve.123133/#post-499779

Two cars are driving at the same constant speed on a straight road, with car 1 in front of car 2. Car 1 suddenly starts to brake with constant acceleration and stops in 10m. At the instant car 1 comes to a stop, car 2 begins to brake with the same acceleration. It comes to a halt just as it reaches the back of car 1. What was the seperation between the cars before they started braking?

All I want to know are enough values to make the same amount equations as there are variables.
Did I miss something or was the question not written properly?

ive attached past attempts.

Thank you
Yes .... you do have enough information to extract the answer. To do this problem properly, you do need to figure out the answer before you write equations. This is a thinking problem - more than anything else. Assume car to be particles.

Now let's assume that car 1 is 'd' meters ahead of car 2.

Car 1 starts to break and stops after 10 m.

Car 2 can travel (d+10) m in that time

For the cars to stop at same "point" (travelling with same speed and acceleration) - car 2 must start to decelerate at the same place on line of travel as car 1 did.

Let us suppose the cars are travelling at speed v initially - distance d apart.

The car 1 starts to decelerate (uniformly) - the time taken (t) to come to stop = 10/(v/2) = 20/v

Uptill now (since the braking of car1 and coming to stop) - the car2 is travelling at constant speed 'v'. How far it would travel in that time (t)?

Now think some more.....
 
Thank you I’ll work on what you said and reply next Sunday (my schedule but I plan to do it). I was worried, see attachment, Quora rejected my post but mentioned there weren’t enough values to solve. I’ll write again soon. Thank you
 

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Yes .... you do have enough information to extract the answer. To do this problem properly, you do need to figure out the answer before you write equations. This is a thinking problem - more than anything else. Assume car to be particles.

Now let's assume that car 1 is 'd' meters ahead of car 2.

Car 1 starts to break and stops after 10 m.

Car 2 can travel (d+10) m in that time

For the cars to stop at same "point" (travelling with same speed and acceleration) - car 2 must start to decelerate at the same place on line of travel as car 1 did.

Let us suppose the cars are travelling at speed v initially - distance d apart.

The car 1 starts to decelerate (uniformly) - the time taken (t) to come to stop = 10/(v/2) = 20/v

Uptill now (since the braking of car1 and coming to stop) - the car2 is travelling at constant speed 'v'. How far it would travel in that time (t)?

Now think some more.....

Still confused but this is what I did.

(i) 0 = V + at

(ii) 10 = 0 + Vt + (a/2)t^2

(iii) 0 = V^2 + 2a10 => a = -(V^2)/20

sub (iii) into (i):

0 = 20V - tV^2 => t = 20/v (iv)

sub (iii) into (ii):

400 = 40Vt - (V^2)(t^2) (v)

sub (iv) into (v):

400 = 800 - (V^2)400/(V^2) which cancels out.

also using your t = 20/v into V = d/t makes d = 20 so the distance between the 2 cars should be 30. If that's the case I can't wrap my head around the linear gradient always ending up twice as much as the parabola's turning point on the y-axis (see attachment)
 

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Still confused but this is what I did.

(i) 0 = V + at

(ii) 10 = 0 + Vt + (a/2)t^2

(iii) 0 = V^2 + 2a10 => a = -(V^2)/20

sub (iii) into (i):

0 = 20V - tV^2 => t = 20/v (iv)

sub (iii) into (ii):

400 = 40Vt - (V^2)(t^2) (v)

sub (iv) into (v):

400 = 800 - (V^2)400/(V^2) which cancels out.

also using your t = 20/v into V = d/t makes d = 20 so the distance between the 2 cars should be 30. If that's the case I can't wrap my head around the linear gradient always ending up twice as much as the parabola's turning point on the y-axis (see attachment)
You stated:

"....also using your t = 20/v into V = d/t makes d = 20 so the distance between the 2 cars should be 30"

No!

I defined 'd' in response#2 to be - "Now let's assume that car 1 is 'd' meters ahead of car 2. "
 
I don’t know where to go from here. What’s the right question to ask you? So 30m apart is not the answer? Any comment about me thinking linear gradient y being double to parabola max y? Also can you atleast tell me yes or no you’ve got an actual number you’ve extracted and not a function due to not enough values given?
 
u1 = u2 = u

a1 = a2 = a

t(s)

T at s = 10m & V = 0m/s

And D = uT + 10

V = at + u => 0 = aT + u => a = -u/T (i)

s = (1/2)at^2 + ut => 0 = aT^2 + 2uT – 20 (ii)

also:

V^2 = u^2 + 2aS => a = -(u^2)/20 (iii)

1st way (i) into (ii):

0 = -u(T^2)/T + 2uT – 20
= -uT + 2uT – 20 => uT = 20

2nd way (i) into (iii):

-u/T = -(u^2)/20 => -20u = -(u^2)T
=> uT = 20

So:

D = uT + 10
= 20 + 10
= 30m

My trouble was trying to solve individual variables u,T,a. There’s not enough info to extract these values, there’s enough to extract the collective variable uT = 20 which is enough to solve the problem D = uT + 10 = 30m the initial distance between the 2 cars. Thanks for helping, I really appreciate it, the clarity is relieving.
 
As the cars have the same constant speed and deceleration we can expect terms to cancel out in any equation. This suggests we could tackle the problem without setting up any equation.
Consider three points on the road - C where each car comes to rest, B where Car 1 starts braking and A the position of Car 2 when Car 1 brakes. Because the two cars have the same initial speed and same deceleration, Car 2 must also start braking at B. Car 2 arrives at B as Car 1 reaches C. So Car 2 travels A to B in the same time as Car 1 travels B to C. In that time Car 2 has a speed twice that of decelerating Car 1 and so covers twice the diatance. As B to C = 10 m, A to B, the original distance between the cars, = 20 m.
 
As the cars have the same constant speed and deceleration we can expect terms to cancel out in any equation. This suggests we could tackle the problem without setting up any equation.
Consider three points on the road - C where each car comes to rest, B where Car 1 starts braking and A the position of Car 2 when Car 1 brakes. Because the two cars have the same initial speed and same deceleration, Car 2 must also start braking at B. Car 2 arrives at B as Car 1 reaches C. So Car 2 travels A to B in the same time as Car 1 travels B to C. In that time Car 2 has a speed twice that of decelerating Car 1 and so covers twice the diatance. As B to C = 10 m, A to B, the original distance between the cars, = 20 m.
Thank you, and how are you able to say car 2 is twice the speed of car 1 decelerating.
 
The average speed during constant deceleration is (initial speed + final speed) / 2. So with final speed 0, half the initial speed.
 
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