acceleration?

Question

A truck covers 40.0 m in 8.5 seconds while smoothly slowing down to final speed 2.80 m/s. (a) find its original speed. (b) find its acceleration

Answer 1

v=velocity t=time x=initial speed
acceleration is (2.8-x)/8.5
v=(2.8-x)/8.5*t+x
integrate v to get distance
int[from t=0 to t=8.5]((2.8-x)/8.5*t+x)dt
(2.8-x)/8.5*t^2/2+x*t -> (2.8-x)/8.5*8.5^2/2+x*8.5
Set the distance = 40m and solve x
(2.8-x)/8.5*8.5^2/2+x*8.5 = 40
x=6.612m/s
get the acceleration from the formula above
(2.8-x)/8.5= -0.448 m/s^2

Answer 2

The formula is distance (s) = 1/2 acceleration * time squared. So 40=1/2a(8.5)^2 or

80=a 72.25

so the rate of acceleration is just barely more than 1 m/sec^2

The original speed is about 11.5m/sec

Answer 3

The average speed is (u+ v) / 2; and is also equal to total distance / time.

The average speed = 40/ 8.5 = 4.71 m/s.

Initial speed is, u = 2 x 4.71 - 2.8 = 6.61 m/s.

Acceleration is {change in speed / time} = (2.8 - 6.61) / 8.5 = - 0.45 m/ s^2.

Negative sign indicates that the speed is decreasing.

Answer 4

speed = distance / time
acceleration= (velocity -velocity0)/ (time- time0)

40.0/8.5 = 4.705882 (speed)

I don't remember how to do acceleration, but this hopefully can help you get started.

Answer 5

It would be wrong to do your homework for you, but I see no reason why we can't point you in the right direction. Google the word "Kinematic". You should find all the information you need to solve your problem.

Answer 6

oh ok--park the truck and climb in the sleeper

Answer 7

Why should we do your homework for you?

Answer 8

Why?