Answer:
14ft/sec
Step-by-step explanation:
GIVEN: An automobile traveling at a rate of 30 ft/sec is approaching an intersection. When the automobile is 120 ft from the intersection, a truck traveling at the rate of 40 ft/sec crosses the intersection.
IN SKETCH BELOW............. IN THE ATTACHMENT
A IS AUTOMOBILE AND V IS TRUCK AT TIME T= 0
TRUCK IS AT M AFTER T SECONDS...AUTOMOBILE IS AT L AFTER T SECS.
AT=120 '
After T seconds
AL= 30T
LV= 120-30T
VM= 40T
LM = [tex]\sqrt{(120-30T)^2+40T^2}[/tex] ( let this be s)
now , ds/dt = [tex]\sqrt{\frac{2(-30)(120-30T)+3200T}{2\sqrt{(120-30T)^2+1600T^2} } }[/tex]
[tex]\frac{ds}{dt} = \frac{2500T-3600}{\sqrt{(120-30T)^2+1600T^2} }[/tex]
now in this equation put T=2 we get
[tex]\frac{5000-3600}{\sqrt{3600+6400} }[/tex]
ds/dt= 1400/100= 14 FPS
hence the two vehicles are separating at 14 ft/sec at T=2 seconds