quick question - resisted motion in a straight line (MECHANICS) (1 Viewer)

[blackops23]

Hi guys, this seems like a relatively simple question, there's just one tiny thing I'm having trouble with:

Q. A particle moves in a straight line against a resistance of magnitude kv per unit mass. At a fixed point O on the line, the initial speed is 50m/s, and the speed is reduced to 5m/s after it has moved 10m.

Find (i) the distance for the speed to reduce to 20m/s.

Here's what I did:

a = -kv
v(dv/dx) = -kv
dv = -k dx
v+c = -kx

v = 50 , x = 0

so v - 50 = -kx

we need to find x when v = 20, so x = 30/k

Question is: HOW DO I FIND THE VALUE OF K?

What i did was chuck in x=10, and v = 5 in the equation v-50=-kx, so i got k=4.5 (which probably is wrong)

Am I allowed to do this? Or is there some other way i have to find k? What is the real value of k?

Thanks guys, appreciate the help.

 

[bleakarcher]

i also got k=4.5. ur right lol. y did u just stop? keep going

 

[blackops23]

so is the answer to (i) 6.6666m?

Also (ii) FInd the time required for speed to reduce to 20m/s -- i got 0.2 seconds is that wrong or right?

 

[bleakarcher]

ur first answer is correct and so ur second should be as well. just use a=dv/dt