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At the start of the journey , there is opposite air resistance and frictional force of 3200 N which causes the truck to move at an acceleration of 3.52 m/s^2 with a resultant force of 8800 N (12000-3200) in the forward direction. As it speeds up , the counter forces (Air resistance and Friction which reduces its acceleration) also increase at the same rate until the resultant force acting on the truck is zero and the then car moves at constant speed.
For example : At the start journey its acceleration was 3.52m/s^2 . After a time of 20 seconds the opposite force acting on the truck is 6000 N so by using F=ma formula ( where F= 12000-6000=6000N , m=2500) acceleration will be 2.4 m/s^2 and this shows that its acceleration has been reduced . And later on when resultant force = 0 , acceleration would be zero and the truck will move at constant speed.
I hope you get it though Im not sure if its right or not
but there is no proportion of forward force to counter forces until the truck is climbing a hill or mountain... it doesnt matter that if the forward force is increasing or not, I guess there is a problem with ur answer there nd i knw hw to find acceleration but couldnt get how the truck will decelerate and then come back to constant speed.