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Particle P released from rest at O. Falls freely under gravity until reaching point A which is $1.25$m below O.
(i) Find speed of P at A and time taken for P to reach A.
P continues to fall, but now its downward acceleration $t$ seconds after passing through A is $(10-0.3t)$ metres per second square.
(ii) Find the total distance P has fallen, $3$ s after being released from O.

I have solved (i) and the speed is $5$ m/s and time is $0.5$ s. (ii) is easy, I think, but the given answer is $44.2$ m, while my answer is coming $43.7$ m.

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  • $\begingroup$ What is your assumption of g for part i)? $\endgroup$ – Doug M Nov 28 '16 at 17:56
  • $\begingroup$ 10, as given to be used. otherwise I would have used 9.81 $\endgroup$ – Hammad Shariq Nov 28 '16 at 17:59
  • $\begingroup$ the answers I have found are given just like part (ii). $\endgroup$ – Hammad Shariq Nov 28 '16 at 17:59
  • $\begingroup$ It looked like you had used 10. But that is vital information to the problem. In order to help. we need to to be clear on the assumptions. $\endgroup$ – Doug M Nov 28 '16 at 18:13
  • $\begingroup$ You are right, I forgot to mention that. Will take care in the future. :) $\endgroup$ – Hammad Shariq Nov 28 '16 at 18:14
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conventionally, down is negative... It would probably be easier to do this all in terms of positive numbers, but I am going to stick with convention.

if $g = -10$ then the time to $A = 0.5 s$ and the velocity at $A$ is $-5 m/s$

$3s$ from $O$ is $2.5 s$ after passing though $A$.

$a = \frac {d^2x}{dt^2}= -10+0.3t\\ v = \frac {dx}{dt}= -5 - 10t + 0.15t^2\\ x = -1.25 - 5t -5t^2 + 0.05 t^3$

$t=2.5$ solve for $x.$ Again $x$ will be negative, (distance below $O$). $|x|$ is the distance from $O.$

I get $44.2$

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  • $\begingroup$ OK, so I was not using the constants from part (i) in the integrals of part (ii), but the rest was easy. Thanks a lot. $\endgroup$ – Hammad Shariq Nov 28 '16 at 18:13

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