help with a motion of a particle question (1 Viewer)

[FOB all the way]

please help


A train runs between 2 stations stopping at each. Its velocity after leaving the first station is given by v=1/5t(4-t) Find:
a) the time taken to travel between the 2 stations
B) the maximum velocity attained
c) the distance between the 2 stations

i can do a but need help with b and c


thanks

 

[FOB all the way]

sorry i just worked out part c but i can not get b

i dont no how to find the maximun velocity

 

[PC]

A train runs between 2 stations stopping at each. Its velocity after leaving the first station is given by v=1/5t(4-t) Find:
a) the time taken to travel between the 2 stations
B) the maximum velocity attained
c) the distance between the 2 stations

Maximum velocity occurs when the acceleration is equal to zero. The particle will be speeding up when the acceleration is positive. When it reaches its maximum, it will then start slowing down. That's cause the acceleration starts acting against it, or when acceleration becomes negative. At some point, the acceleration changes from positive to negative, which is zero.

V = 1 / 5t(4-t)
= [5t(4-t)]^-1
Therefore:
A = -1 x [5t.-1 + (4-t).5] x [5t(4-t)]^-2
= -1 x [-5t + 20 - 5t] x [5t(4-t)]^-2
= -1 x [20 - 10t] x [5t(4-t)]^-2
= [10t - 20] x [5t(4-t)]^-2
= 10[t - 2][5t(4-t)]^-2
= 10(t - 2) / [5t(4-t)]^-2

Now when A = 0:
10(t - 2) = 0
t - 2 = 0
t = 2

SO maximum velocity = 1 / 5(2)(4-2)
= 1 / 5 x 2 x 2
= 1/20 m/s

Hope this helps.