A flatbed truck with a box sitting on the bed accelerates from rest on level ground. The truck’s speed as a

function of time is 15t –2t^2 during this acceleration. The truck reaches its maximum speed and cruises along at this speed for 4 minutes. It then slows to a stop with an acceleration of 2 m/s/s.
a) What is the net force as a function of time, acting on the truck during the speeding up portion?
b) What is the maximum speed the truck attains?
c) At what time does the truck reach maximum speed?
d) How far does the truck travel during the entire trip?
e) How much work is done on the 588 N box in the bed of the truck while speeding up?
f) What force does that work on the box?
g) How much total work is done on the box during the entire trip?

v=15t-2t²

a=dv/dt = 15 - 4t.
(a)F=m•a=m• (15-4t),

For finding v(max):
a=15 - 4t=0
15=4t
(c) t=15/4 (s)

(b)v(max)=15t-2t² =
=15•15/4 -2•(15/4)²=28.125 m/s.

(d) s1=∫(15t-2t²)dt=15t²/2 – 2t³/3 =
=15•15²/2•4² -2•15³/3•4³ = 70.31 m.
s2 = v(max) •t=28.125•4•60=6750 m.
s3= v²/2a=(28.125)²/2•2 = 197.8 m

s1+s2+s3 = 70.31+6750+197.8 =7018 m.

(e) The work of the net force during the sppeding up portion
W1=ΔKE=mv²/2=(588/9.8) •28.125²/2=2.37•10^4 J.
(f) F=m•a=m• (15-4t),
(g) W2=0 (since the net force is zero),
W3=ΔKE =0- mv²/2=0-(588/9.8) •28.125²/2=
=0-2.37•10^4 J= - 2.37•10^4 J.
W=W1+W2+W3=0

To answer these questions, we need to use the equations of motion and the principles of Newtonian mechanics. Here's how we can tackle each question:

a) The net force acting on the truck is given by Newton's second law: F = ma. In this case, the acceleration is given by a(t) = 15 - 4t (since the truck is slowing down at 2 m/s^2). Therefore, the net force can be calculated by multiplying the mass of the truck by the acceleration.

b) The maximum speed attained by the truck is the maximum value of its speed function. To find this maximum, we can differentiate the speed function with respect to time and set it equal to zero. By solving this equation, we can determine the time at which the truck reaches maximum speed. Plugging this time into the speed function will give us the maximum speed.

c) To find the time at which the truck reaches maximum speed, we need to solve the equation obtained in part b.

d) The distance traveled by the truck during the entire trip can be found by integrating the speed function over the entire time interval. The integral of the speed function gives the displacement function, which represents the position of the truck with respect to time.

e) The work done on an object is given by the force acting on it multiplied by the distance it moves in the direction of the force. In this case, the force can be calculated using Newton's second law and the acceleration function found in part a. The work done on the box is then given by multiplying the force by the distance traveled by the truck.

f) Since the box is contained within the truck, the force exerted on the box will be the same as the net force acting on the truck found in part a.

g) To find the total work done on the box during the entire trip, we need to integrate the force function over the entire time interval and multiply it by the displacement of the truck.

By following these steps, we can find the answers to all the given questions.