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From the acceleration of a particle, that is given as $a = 2t + 4$, I have worked out velocity from integration as $V = t² + 4t + c$. The displacement of the object is 30 m after 3 s, how do I find $c$? I am also unsure what is meant by finding displacement in terms of time?

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    $\begingroup$ You’ve been able to find velocity in terms of time. What’s confusing you about finding displacement in terms of time? $\endgroup$ – amd Apr 21 '20 at 19:02
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This seems like question in physics site, but this is simple, so whatever.

Your velocity at time 't' was derived by integrating acceleration at time 't' and constant 'c' was added as we don't know initial velocity, since acceleration only talks about change in velocity.

Here c is velocity at time t=0.

Similarly by integrating velocity, we get location at time 't' as

$X = \dfrac{t^3}{3}+2t^2+ct+k$

Here k is location at time t=0. In other words, initial location.

Displacement is net change in location of an object. Assuming our first point as "t=0"

So, displacement will be $Displacement =\dfrac{t^3}{3}+2t^2+ct$

since t is given as $3s$, Displacement is equal to $9+18+3c = 27 +3c$

Equating it with $30 m$ will give $c=1$

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  • $\begingroup$ Thank you, i worked out that the constant was 3, which was marked correct, im an engineer doing a Hnc in college, maths and especially calculus was never my strong point. $\endgroup$ – Daniel Jordan Apr 22 '20 at 16:07
  • $\begingroup$ Cheers from a fellow engineer. $\endgroup$ – xax May 15 '20 at 11:29

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