The acceleration of a particle at time #t# seconds is given by # a = 6t#. Given that #v=10# when #t=0# and that #x=7# when #t=0# find #x# when #t=2#?

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Answer 1 · 2017-02-18T22:34:34

The position #x# at time #t# is given by;

# x = t^3 + 10t + 7 #

And so when #t=2# we have:

# x = 35#

Acceleration (#a#) is defined as the rate of change of velocity (#v#) wrt time (#t#). thus

# a = (dv)/dt => (dv)/dt = 6t#

This is a First Order separable Differential Equation which we can just integrate to get:

# v = 3t^2 + A #

Using the initial condition #v=10# when #t=0# we get:

# \ \ \ 10 = 0 + A => A=10 #
# :. v = 3t^2 + 10 #

Velocity (#v#) is defined as the rate of change of position (#x#) wrt time (#t#). thus

# v = (dx)/dt => (dx)/dt = 3t^2 + 10#

Again this is a First Order separable Differential Equation which we can just integrate to get:

# x = t^3 + 10t + B #

And using the initial condition #x=7# when #t=0# we get:

# 7 = 0 + 0 + B => B= 7#

Hence the position #x# at time #t# is given by;

# x = t^3 + 10t + 7 #

And so when #t=2# we have:

# x = 2^3 + 10*2 + 7 =35#