Maximum distance ahead of trooper reached by red car

[Rijad Hadzic]

Homework Statement

A trooper is moving due south along the freeway at a velocity speed of 21 m/s. At time t= 0, a red car passes the trooper. The red car moves with constant velocity 28 m/s. At the instant the troopers car is passed, the trooper begins to speed up at a constant rate of 2 m/s^2. What is the maximum distance ahead of the trooper that is reached by the red car?

Homework Equations

[itex] V_ox + a_xt = V_x
\Delta x = (1/2) (V_{ox} + V_x)(t)
\Delta x = V_{ox}t + (1/2)a_xt^2 [/itex]

The Attempt at a Solution

Ok so I attempted to make a graph where I manually calculated the difference.

my algorithm for troopers meters from origin goes as follows:
use [itex] V_ox + a_xt = V_x [/itex] to calculate V_x, then use

[itex]\Delta x = (1/2) (V_{ox} + V_x)(t) [/itex] to calculate meters from origin

so for example, for 2 seconds after start, I would use

[itex] V_ox + a_xt = V_x [/itex] and plug in: [itex] 21 m/s + 2 m/s^2 (2 s) = V_x = 25 m/s [/itex]

now using 25 m/s for V_x use [itex]\Delta x = (1/2) (V_{ox} + V_x)(t) [/itex] to calculate meters from origin..

[itex] \Delta x = (1/2) (25 m/s + 21 m/s) (2 s) = 46 m [/itex]

using this same formula for the rest of my graph, my graph goes as follows:

r.c = red car's meters from origin, t = trooper's meters from origin d = difference
1 s, rc= 28m, t =22m, d =6m
2 s, rc = 56m, t= 46m, d= 10 m
3 s, rc = 84m, t= 72m, d = 12 m
4 s, rc = 112m, t = 100m, d= 12m
5 s, rc = 140m, t = 130m, d = 10 m

I saw the difference already got smaller after 5s so I calculated 3.5 s because its between 3 and 4s
3.5 s, rc = 98 m, t = 85.75 m, d = 12.25 m

so my answer is the max distance the red car will be away from the trooper will be 12.25 m

Now using another method,

[itex] \Delta x_{r.c} = 28 x
\Delta x_{t} = (1/2)(42x+2x^2)
difference formula = 7x-x^2 [/itex]

I take the derivative of difference formula and = to zero, this gives me a time of 7/2 seconds. Plugging back into the difference formula, and I get 12.25 m, I put 7/2 seconds in [itex] \Delta x_{r.c} and \Delta x_t [/itex] and subtract and still get 12.25 m,

BUT my book says the answer is 16 meters.

I can see why I got the same answer using both methods, because I used the same kinematic equations, but maybe I'm not suppose to do it this way? Does anyone see where I failed?

 

[cnh1995]

I think your answer is correct. I don't see any mistake in your working.

 

[Rijad Hadzic]

I think your answer is correct. I don't see any mistake in your working.

Thanks for the reassurance mate.. just looking for another response or two to make sure I didn't overlook anything..

 

[haruspex]

Thanks for the reassurance mate.. just looking for another response or two to make sure I didn't overlook anything..

I confirm 12.25m, but would like to show you an easier way.
Thinking of it in the red car's reference frame, at time zero the police car was moving away from it at 7m/s, decelerating at 2m/s². The max distance will be when that relative speed is zero.
Using the equation v²=u²+2as, 7²=0+2(2)s, s=12.25m.

 

[Rijad Hadzic]

I confirm 12.25m, but would like to show you an easier way.
Thinking of it in the red car's reference frame, at time zero the police car was moving away from it at 7m/s, decelerating at 2m/s². The max distance will be when that relative speed is zero.
Using the equation v²=u²+2as, 7²=0+2(2)s, s=12.25m.

Thanks man. You always know how to make things simple as possible.

 

[haruspex]

Thanks man. You always know how to make things simple as possible.

By the way, 16m would have been the right answer if the initial speeds had differed by 8m/s instead of 7.

 

[Rijad Hadzic]

By the way, 16m would have been the right answer if the initial speeds had differed by 8m/s instead of 7.

So do you still think the answer is 12.25 m ?

The values are correct, 21 m/s, and 28 m/s.

The problem is making me think that when they say "passes," they mean he was already a head an x amount of miles?

 

[haruspex]

So do you still think the answer is 12.25 m ?

The answer to the problem as given, yes.

The values are correct, 21 m/s, and 28 m/s.

Ok, thanks for checking.

The problem is making me think that when they say "passes," they mean he was already a head an x amount of miles?

No, it is definitely a mistake in the book. I think it was intended to specify either 20m/s and 28m/s, or 21m/s and 29m/s.