[SOLVED] Simple Mechanics Integration

Sep 2009
148
2
Q) A particle P is held at rest at a fixed point O and then released. P falls freely under gravity until it reaches the point A which is 1.25m below O.

(i) Find the speed of P at A and the time taken for P to reach A.

The particle continues to fall, but now its downward acceleration t seconds after passing through A is (10 − 0.3t)ms^-2

(ii) Find the total distance P has fallen, 3 s after being released from O.

I've solved part one to get...

t=0.5s and v=5ms^-1

But I'm unable to do part (ii). Don't know where to start and how to solve this. I've tried to integrate dv/dt but not getting anywhere.

Thanks!
 
Sep 2008
1,261
539
West Malaysia
Q) A particle P is held at rest at a fixed point O and then released. P falls freely under gravity until it reaches the point A which is 1.25m below O.

(i) Find the speed of P at A and the time taken for P to reach A.

The particle continues to fall, but now its downward acceleration t seconds after passing through A is (10 − 0.3t)ms^-2

(ii) Find the total distance P has fallen, 3 s after being released from O.

I've solved part one to get...

t=0.5s and v=5ms^-1

But I'm unable to do part (ii). Don't know where to start and how to solve this. I've tried to integrate dv/dt but not getting anywhere.

Thanks!
for part (2) , i think calculus is not necessary here , instead just use the motion formulas, ie s=ut+1/2 at^2

For the first 0.5 s , it has travelled 1.25 m

For the next 2.5 s ,

t=2.5 s , u=final speed of particle at A , 5 m/s and a is given

s=5(2.5)+1/2 (10-(0.3)(2.5))(2.5)^2
 
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Sep 2009
148
2
"s=5(2.5)+1/2 (10-(0.3)(2.5))(2.5)^2"

What have you done here, can you please tell me?
The answer is not coming out right when I solved the above and added 1.25m to it.

The correct answer is 44.2m

Btw can you also try this with calculus and show me your steps? Thanks!

Edited....

I have a test coming up and need to get a grasp on this question. So, kindly someone help me with this.

Thanks!
 
Last edited:

HallsofIvy

MHF Helper
Apr 2005
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for part (2) , i think calculus is not necessary here , instead just use the motion formulas, ie s=ut+1/2 at^2
No, you can't. That formula (which is derived using Calculus!) assumes a constant acceleration, which is not true here.
With an acceleration of a= 10- .3t, its speed at any time t after the initial drop is \(\displaystyle v= 10(t- .5)- .15(t- .5)^2+ 5 m/s\). To find the distance fallen in 3 seconds, integrate that again and evaluate at t= 3.