Acceleration + distance

[Ings19]

A mass of 40tonnes is accelerated from rest for 20s then runs uniformly for 5 mins and is finally brought to rest. Assuming acceleration and deceleration are 0.7m/s2

calculate;

a) total distance travelled)

b)total force required during this period (ignoring friction)

for a I have calculated 4480m

for b I have calculated 28kn. (F=m x a = 40000 x0.7)

[FizziksGuy]

For part a, I would use kinematics: d=vi*t+0.5*a*t^2, and then 300s*vf, and then the deceleration distance.

For part b, the total force required, F=ma looks good, but make sure your mass is in kg.

[Ings19]

Thanks, for part A I used;

final velocity= initial velocity+ rate if acceleration x time = 14m/s

Distance during acceleration and deceleration = (14/2) x 20s = 140m

Distance= uniform velocity x time taken = 14x300s= 4200m

so answer to part a come to 140+4200+140=4480m

then part b 40tonnes =40,000kg

F=mxa = 40,0000x0.7 =28,000n =28Kn

[Ings19]

What do you guys think?

[FizziksGuy]

I think you've got it.