how would the following X versus T graph look like?

Question

At t= 0s a truck is stopped at a traffic light. When the light turns green, the truck starts to speed up, and gains speed at a constant rate until it reaches a speed of 20m/s 8 seconds after the light turns green. The truck continues at a constant speed for 60m . Then the truck driver sees a red light up ahead at the next intersection, and starts slowing down at a constant rate. The truck stops at the red light, 180m from where it was at .


Draw accurate graph for the motion of the truck.

I understand that the first part of the graph is like a parabola, and at one point its linear, and then becomes almost horizontal..can anyone draw this graph please, or the values or x between interval time of 0-15

Answer 1

Here are the equations which you can graph
from t=0 to t=8
v(t)=a*t
20=a*8
a=20/8
a=5/2
so
v(t)=5*t/2
and
d(t)=.5*5*t^2/2
simplify
d(t)=5*t^2/4
That is your parabola.
The truck travels a distance of 80 m

Then for t=8 to t=? for 60 m
How long does it take to travel 60 m at 20 m/s?
3 seconds

The gragh is
d(t)=80+20*(t-8) for t=>8 and t<=11

So far the truck has traveled 140 meters

the final leg sounds like a constant deceleration for 40 meters.
v(t)=20-a(t-11) for t>=11 until v(t)=0
let's solve for t when the truck stops
0=20-a*(t-11)
or
20=a*(t-11)
since we also know the distance
d(t)=140+20*(t-11)-
.5*a*(t-11)^2 for t>=11 until d(t)=180

when d(t)=180, v(t)=0, so
180=140+20*(t-11)-
.5*a*(t-11)^2
plug in 20=a*(t-11)
180=140+20*(t-11)-
.5*20*(t-11)

solve for t and graph the last bit.
t=15. So the last segment spans 4 seconds

now solve for a
a=5 (positive since I used a negative in the equation)

d(t)=140+20*(t-11)-
.5*5*(t-11)^2 for t>=11 t<=15

Here's the graph
http://i142.photobucket.com/albums/r88/odu83/Truck.jpg


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