Determining velocity from acceleration and distance?
An x-ray tube accelerates electrons from rest at 3.0×1014 m/s^2 through a distance of 12 cm
Find the electrons'velocity after this acceleration
Find the acceleration time
i am confused as where to start this problem.
- 7 years agoFavourite answer
v^2 = u^2 + 2as
v^2= 0^2 + 2*3*10^14*0.012
v^2 = 7.2*10^12
v = 2.68*10^6
- oldprofLv 77 years ago
When in doubt start with the basic distance = average speed X time; S = Vavg T, where S = .12 m is given. Also given is A = 3E14 m/s^2 the acceleration and we are looking for V the velocity after accelerating over the distance.
Vavg = (U + V)/2 when A is constant. U = 0; so Vavg = V/2. Now we have S = V/2 T so V = 2S/T = ? and we need to find T. V = U + AT = AT so that T = V/A and that gives us V = 2S/T = 2SA/V or V^2 = 2AS and V = sqrt(2AS) which is what the other answer gave you, but now you know the physics where that equation came from. Both A and S are given; plug and chug.
But remember this...when "confused as where to start" start with good old S = Vavg T and derive what you need from it.