Question
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.
f=ma
f=1,188(9.8)
f=11642.4N
This looks wrong to me. Can someone explain this to me.
Answers
bobpursley
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.
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.
Anonymous
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
Kat
What is the average force need to stop a 1500 N car in 8.0 S if it is traveling at 90 km/hr?