[المتفيزق]

A 2000-kg truck traveling at a speed of 6.0 m/s makes a 90° turn in a time of 4.0 s and emerges from this turn with a speed of 4.0 m/s. What is the magnitude of the average resultant force on the truck during this turn?
a. 4.0 kN b. 5.0 kN c. 3.6 kN d. 6.4 kN e. 0.67 kN
ok...
here we want to calculate the average accelleration ...
now notice that we may let the truck move first along x axis and then moves 90 deg. to y axis so that :
vo=6i m/s v =4j m/s and delta t =4s
this means that:
average acc. = (4j - 6i)/4 m/s2
which has a magnitude of sqrt( 16+36)/4 = 7.2/4 m/s2 nearly
this means that the force exerted on the truck is F=ma = 2000*7.2/4=3600N
or 3.6kN
and again... thanks