Question

I need help with this question with explanation. Thanks
A particle of mass m starts from rest at position x = 0 and time t = 0. It moves along the positive x-axis under the influence of a single force Fx = bt, where b is a constant. The velocity v of the particle is given by
A) bt/m
B) bt2/2m
C) bt2/m
D) b√t/m
E) b/mt

Answer 1

this is analogous to the relationship between
... acceleration, velocity, and distance

acceleration is velocity changing with time ... v = a t
... and velocity is distance changing with time ... d = v t
... so , d = 1/2 a t^2

in this case
... force is acceleration changing with time ... a = (b/m) t
... and acceleration is velocity changing with time ... v = a t
... so , v = 1/2 (b/m) t^2

Answer 2

Hello, I was looking at this problem and probably can explain what is going on

so what you did is you tried to substitute acceleration for change in velocity/ change in time and got (bt^2/M) just like me, well you only found change of velocity, NOT velocity

your answer should be B

F= bt
ma=bt //replace Force with ma
a=bt/m // get a by itself
v=bt^2/2m // take the integral of both sides to find your velocity and answer (remember b and m are constants)

I hope this helped

Answer 3

Hello, I was looking at this problem and probably can explain what is going on

so what you did is you tried to substitute acceleration for change in velocity/ change in time and got (bt/M) just like me, well you only found change of velocity, NOT velocity

your answer is bt^2/2m (which is NOT listed above)

F= bt
ma=bt //replace Force with ma
a=bt/m // get a by itself
v=bt^2/2m // take the integral of both sides to find your velocity and answer (remember b and m are constants)

I hope this helped

Answer 4

F = bt
ma = bt
a = (bt)/m

Note that you can not use the standard kinematics equations because the acceleration is not constant! The acceleration is a function of time; you need to use an integral.

∫(bt)/m dt - acceleration times time, but the acceleration isn't constant.
= (bt^2)/(2m)