110 km/hr(1000/3600) = 30.6 m/s
119 km/hr = 33.1 m/s
x position of truck = -211.7+ 30.6 t
v car = 2.3 t until v = 33.1
at t = 14.4 s and x = (1/2)(2.3)(14.4^2)
= 238 m
after that xcar = 238 + 33.1 (t-14.4)
so during car acceleration period:
d = Xcar - Xtruck = .5(2.3)t^2 - 211.7 + 30.6 t
or
d = 1.15 t^2 + 30.6 t - 211.7
does that have a minimum between t = 0 and t = 14.4?
I do not know if you do calculus, if not you will have to find the vertex of that parabola with completing the square
d(d)/dt = 0 at min = 2.3 t +30.6
t = 30.6/2.3 = 13.3 seconds to minimum d
then
d = 1.15(13.3^2) +30.6 t -211.7
= 22.3 meters
To save fuel, some truck drivers try to maintain a constant speed when possible. A truck traveling at 110.0 km/hr approaches a car stopped at the red light. When the truck is 211.7 meters from the car the light turns green and the car immediately begins to accelerate at 2.30 m/s2 to a final speed of 119.0 km/hr. How close does the truck come to the car assuming the truck does not slow down?
1 answer