To save fuel, some truck drivers try to maintain a constant speed when possible.

Question

1. To save fuel, some truck drivers try to maintain a constant speed when possible. A truck traveling at 83.0 km/hr approaches a car stopped at the red light. When the truck is 74.3 meters from the car the light turns green and the car immediately begins to accelerate at 3.90 m/s2 to a final speed of 114.0 km/hr. How close does the truck come to the car assuming the truck does not slow down?
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Explanation / Answer

To save fuel, some truck drivers try to maintain a constant speed when possible. A truck traveling at 83.0 km/hr approaches a car stopped at the red light. When the truck is 74.3 meters from the car the light turns green and the car immediately begins to accelerate at 3.90 m/s2 to a final speed of 114.0 km/hr. How close does the truck come to the car assuming the truck does not slow down?

INITIAL DISTANCE BETWEEN CAR AND TRUCK = 74.3 M ......

TAKING TRUCK POSITION AS START POINT ..ZERO DISTANCE .

LET THIS BE TIME ZERO ...SO ....FOR THE CAR MOTION WE GET IN TIME T SECS ..

SPEED OF CAR =V = 3.9*T M/SEC. .....1................

FOR T < 114/(3.9*3.6) SEC........

AND V=114 ................................................2....................

FOR T>114/(3.9*3.6) SEC........................................

SINCE FINAL SPEED OF CAR IS 114 KMPH > 83 KMPH OF TRUCK , WE NEED CONSIDER ONLY THE TIME TILL CAR REACHES ITS CONSTANT SPEED ..

HENCE AT ANY TIME T .. FOR T<114/(3.9*3.6)=8.12 SEC...WE GET ....

S=74.3+ 0.5*3.9*T*T.......................................................3...........................

THE TRUCK WOULD BE AT.... 83T /3.6=23.056T....

THE RELATIVE GAP = X = 74.3+ 0.5*3.9*T*T - 23.056 T ........................................4

THE OPTIMUM WOULD BE WHEN ....

DX/DT = 3.9T - 23.056 =0 .....T=5.9117 SEC...

AT THAT TIME WE GET ...

X= 74.3+ 0.5*3.9*5.9117^2 - 23.056 *5.9117 = 142.45 M ........ANSWER....

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