The acceleration of a particle is defined by the relation A=-KV^2 where A is exp

Question

The acceleration of a particle is defined by the relation A=-KV^2 where A is expressed in ft/s^2 and V in ft/s. The particle starts a X=0 with a velocity of 20 ft/s and when X=100 ft the velocity is found to be 15 ft/s. Determine the distance the particle will travel (1) Before the velocity drops to 10 ft/s (2) Before it comes to rest. There is a similar example on Chegg, but the solution skips multiple steps and is very hard to follow. I'm looking to see how the solution is solved. Thanks in advance to all who help me.

Explanation / Answer

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a = V.dV/dS = - K.V^2

So, dV/V = - K.dS

so, ln V = -K.S + C

where C = constant

now at X = 0, V = 20 ft/s

so, ln 20 = - K.0 + C

so C = ln 20

Now, at X = 100 ft, V = 15 ft/s

so, ln 15 = - K . 100 + ln 20

so K = (ln 20 - ln 15)/100

so K = ln (20/15) / 100

so , K = 0.00287

1) so, ln V = - 0.00287 .S + ln 20

so, at V = 10 ft/s , ln 10 = - 0.00287 .S + ln 20

so, S = 241.51 ft

2) at V = 0 ft/s,

ln 0 = - 0.00287 .S + ln 20

so , S = 1043.81 ft

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