F(fwd) - F(bkwd) = M a
Based upon the speed change in 9.0 s, a is actually 2.98 m/s^2, not 3.0. I will use 3.0 anyway. They gave you more information that you need.
Solve for F(fwd)
F(fwd) = 280 + (1354)*(3.0) N
A certain sports car accelerates from 0 to 60 mph in 9.0 s (average acceleration = 3.0
m/s2). The mass of the car is 1354 kg. The average backward force due to air drag during
acceleration is 280 N. Find the forward force required to give the car this acceleration.
4 answers
according to newton's second law;it states that
F=ma
therfore,F(force of car)-F(force of airdrag)=ma
ie:F-F=Ma
inorder to find the "F" transpond one F from left hand side to right hand side.
therfore,F=F+MA
ie:F=280+(1354)x(3.0)
F=4902.
Therfore the force needed to give the car this accelaration=4902.
F=ma
therfore,F(force of car)-F(force of airdrag)=ma
ie:F-F=Ma
inorder to find the "F" transpond one F from left hand side to right hand side.
therfore,F=F+MA
ie:F=280+(1354)x(3.0)
F=4902.
Therfore the force needed to give the car this accelaration=4902.
Not be that person or anything... but your final calculation is incorrect... it is 4342.
my teacher book is the answer is 4.4e3