- #1
Rijad Hadzic
- 321
- 20
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?