Physical Apps of Calculus Question (1 Viewer)

[boredofstudiesuser1]

Coroneos 2U Set 17E Q18

(i) A particle is moving along the x-axis. At time t, it is distant x from 0 and moving with velocity v=dx/dt
Show that acceleration dv/dt = v(dv/dx)

(ii) The acceleration of a body moving in a straight line is proportional to the velocity, i.e. dv/dt = kv, where k is constant. When x=0, v=7 and when x=3, v=13.
Find the value of v when x=2 (Hint: use the result in (i))

 

[InteGrand]

Coroneos 2U Set 17E Q18

(i) A particle is moving along the x-axis. At time t, it is distant x from 0 and moving with velocity v=dx/dt
Show that acceleration dv/dt = v(dv/dx)

(ii) The acceleration of a body moving in a straight line is proportional to the velocity, i.e. dv/dt = kv, where k is constant. When x=0, v=7 and when x=3, v=13.
Find the value of v when x=2

(Hint: use the result in (i))

i) We have

dv/dt = (dv/dx)*(dx/dt) (chain rule)

= (dv/dx)*v = v*(dv/dx) (as dx/dt = v).

ii) We have dv/dt = kv, so from the first part, v*dv/dx = kv.

It follows that dv/dx = k, so v = kx + C for some constant C.

When x = 0, v = 7, so C = 7. So v = kx + 7. When x = 3, v = 13, so 13 = 3k + 7 ==> k = 2.

So v = 2x + 7. When x = 2, v = 11.

 

[boredofstudiesuser1]

i) We have

dv/dt = (dv/dx)*(dx/dt) (chain rule)

= (dv/dx)*v = v*(dv/dx) (as dx/dt = v).

ii) We have dv/dt = kv, so from the first part, v*dv/dx = kv.

It follows that dv/dx = k, so v = kx + C for some constant C.

When x = 0, v = 7, so C = 7. So v = kx + 7. When x = 3, v = 13, so 13 = 3k + 7 ==> k = 2.

So v = 2x + 7. When x = 2, v = 11.

Thank you.