Yes, force = rate of change of momentum
= m a if mass is constant
a = (v2-v1)/t
Remember the parentheses.
Hello,
If I need to find out how long it takes for an object acted on by a force takes to change speed by 1.0m/s^-1 and I know the magnitude of the force can I use the following equation?
a=v2-v1/t in order to establish the time taken?
4 answers
So if the acceleration was 60000 and I wanted to solve how many hours this would take I would:
a=(60000-0)/3600
Which is 16.2 hours? I’m alittle confused as this is how someone has shown me but I don’t understand why the time taken would be a (acceleration) rather than t?
Sorry I know it’s an odd question.
a=(60000-0)/3600
Which is 16.2 hours? I’m alittle confused as this is how someone has shown me but I don’t understand why the time taken would be a (acceleration) rather than t?
Sorry I know it’s an odd question.
Math is kwel
Anon, you would use the mathematical expression which is got from Newton's second law of motion ie F=ma where F is the force acting on the object, m is the mass of the object and a is the acceleration of the object.This will help you to get the acceleration which you have said that =60000 m/s^2. You will then use your equation of:
a=(v2-v1)/t where (v2-v1)=1.0 m/s but not 1.o m/s^-1 as you had written in your question.ie, from a=(v2-v1)/t, it implies that 60000= 1/t, multiplying both sides by t: 60000 t =1, then divide both sides by 60000, ie (60000 t)/60000=1/60000: then t=1.67x10^-5 seconds =((1.67x10^-5)/3600) hours
a=(v2-v1)/t where (v2-v1)=1.0 m/s but not 1.o m/s^-1 as you had written in your question.ie, from a=(v2-v1)/t, it implies that 60000= 1/t, multiplying both sides by t: 60000 t =1, then divide both sides by 60000, ie (60000 t)/60000=1/60000: then t=1.67x10^-5 seconds =((1.67x10^-5)/3600) hours
1 answer