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.
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 $44.2$