Why did you choose 9.8 as the acceleration? It is not falling.
its average velocity during stopping is 89/2 km/hr (change that to m/s)
So it traveled in that time a distance of avgvelocity*9s...
force*distance=initial KE= 1/2 m vi^2
change vi to m/s, then solve for force.
What average force in N is required to stop an 1,188 kg car in 9 s if it is initially traveling at 89 km/h?
f=ma
f=1,188(9.8)
f=11642.4N
This looks wrong to me. Can someone explain this to me.
3 answers
Firstly F=ma but a=(v-u)/t. First convert the 89 Km/h into m/s.
89x1000/3600=24.72m/s
Then using F=m(v-u)/t
F=1188(24.72-0)/9
F=3263.04N
89x1000/3600=24.72m/s
Then using F=m(v-u)/t
F=1188(24.72-0)/9
F=3263.04N
What is the average force need to stop a 1500 N car in 8.0 S if it is traveling at 90 km/hr?
1 answer