The relation between distance #s# and speed #v# is given by #v = (150s)/(3+s)#. Find the acceleration in terms of #s#. Help!?

2 Answers
Steve M
Aug 13, 2017

# a = (67500s)/(3+s)^3 #

Explanation:

We have:

# v = (150s)/(3+s) # ..... [A}

Velocity is defined as the rate of change of the distance wrt time, acceleration is defined as the rate of change of speed wrt time.

# v = (ds)/dt; \ \ \ a=(dv)/(dt) = (d^2s)/(dt^2) #

Differentiating] wrt #t# whilst applying the quotient rule we have:

# (dv)/dt = ( (3+s)(d/dt (150s)) - (150s)(d/dt (3+s)) ) / (3+s)^2 #

# :. a = ( (3+s)( 150(ds)/dt) - (150s)((ds)/dt) ) / (3+s)^2 #

# \ \ \ \ \ \ \ = ( (3+s) 150v - 150sv ) / (3+s)^2 #

# \ \ \ \ \ \ \ = 150v \ ( (3+s) - s ) / (3+s)^2 #

# \ \ \ \ \ \ \ = 150v \ ( (3) ) / (3+s)^2 #

# \ \ \ \ \ \ \ = ( 450v ) / (3+s)^2 #

And substituting for #v# from [A] we have:

# a = ( 450 * (150s)/(3+s) ) / (3+s)^2 #

# \ \ \ \ \ \ \ = (67500s)/(3+s)^3 #

Cesareo R.
Aug 13, 2017

See below.

Explanation:

We have

#(s+3)v-150s = 0# and deriving regarding #t#

#s' v+(s+3)v'-150s'=0# but #s'=v# and #v'=a# so

#v^2-150v +(s+3)a = 0# or

#a = (v(150-v))/(s+3) # and substituting #v# we have

#a = (67500s)/(s+3)^3#

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