Hyperlearning physics - arithmatic killing me

[Addallat]

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I'm getting killed on my timing when it comes down to doing physics problems due to basic arithmetic. it's incredibly frustrating and I wind up taking 4 minutes for a straight forward discrete for example:

Question 46 in Hyperlearning WB:


A rock is dropped from a 128-ft cliff. How long does it take to reach the ground? (Ignore air resistance and take g= 32 ft/sec^2)


A. 2.0
B. 2.8
C. 4.0
D. 5.6

the way I went about solving it:

I recognize that i need to solve for Time (t)
recognized that displacement (d) = 128 ft down in the positive direction
acceleration = 32 down in the positive direction
assumed initial velocity (Vo) is 0

so immediately wrote down the formula d = Vo(t) + 1/2 a(t^2)
simplifying to d = 1/2 a (t^2)
rearranging to solve for t:


t = SQRT (2(d)/a)


it takes me about a minute to get to the above formula (i basically write down my entire thought process on paper...)
and I wind up taking 3 minutes to solve the remainder of the problem due to the math....

I manage to do the arithmetic faster with rounding but can someone tell me if my mindset is right for rounding

I round 128 up to 130 multiple by 2 to get 260 and round the acceleration down by 2 to get 30
260/30 ~~~> which then I quickly figure in my head 30 * 9 = 270 so close enough to 260 then take the square root of 9 to get and answer of 3 and choose B. 2.8 as my answer because it's closest to my answer of 3

Can someone out there please write down their thought process when working through the above problem; I'd just like to see how others solve it




ALSO: Is the arithmetic found in the problem typical for actual problems seen on the MCAT?

 

[Addallat]

Laying in bed thinking about this problem and did it in my head using factoring. Anyone else think factoring is the most efficient way to go about the arithmetic at the end???

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[enervate]

A rock is dropped from a 128-ft cliff. How long does it take to reach the ground? (Ignore air resistance and take g= 32 ft/sec^2)

t = SQRT (2(d)/a)

I don't need to write down y = v_0*t + (1/2)*a*t^2 to arrive at y = (1/2)*a*t^2 when there is no initial velocity. Even if I did, I can just scratch out v_0*t since it's null anyway.

From there, plug in numbers. I already know that 128 is a multiple of 32: 128/32 = 4. Remember that.

Plug those numbers into the formula to get 128 = (1/2)*32*t^2. 128/32 is 4 so that means t^2 = 4*2 and t = sqrt(8).

Hell if I know what the square root of 8 is but I know that it's more than sqrt(4) = 2 and less than sqrt(9) = 3.

Looks like my only option is 2.8. Go with it.



How easy is that?





edit: I guess sqrt(8) = sqrt(4) * sqrt*(2) = 2 * sqrt(2). Since sqrt(2) is one of those things that I've memorized for trigonometry, that winds up being 2 * 1.41 = 2.82.

 

[Addallat]

Ahhhh that was a great response thank you!

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[enervate]

Glad it was helpful.

 

[catzzz88]

The way I like to do these is by TBR "short cut" method. You might find your time cut by 1/3 with this method.

1) Convert feet to meters --> 125(0.3) = (125*3)/10 = 375/10 = 37.5
The rock falls 37.5 meters. How long does that take?

2) Use equation delta(h) = 5t^2
--> 37.5 = 5t^2
Since you are taking the square root, you should notice that it is critical to round up in this case--otherwise you would be solving for sqrt(7)... yikes!
t = sqrt(37.5/5) = sqrt(8) = sqrt(4*2) = 2*sqrt(2) = 2*1.4 = 2.8
(The above method for solving sqrt is called factoring and it is really easy and really fun! I would recommend practicing this a bit as you can go up to very high numbers using this method! Impress your friends and get a girlfriend by showing her how you can solve for the sqrt(15932) haha)

Love this shortcut, you might want to consider memorizing this formula as it is extremely useful to do these questions quickly. However, this question would have been impossible to do by this method without knowing the conversion from feet to meters (you can just remember 1/3).

The other useful kinematics shortcut equation that I like to use is v(final) = sqrt(2gh)

 

[tchoudha]

x=vit+1/2gt^2

vi=0 at the apex
x=1/2gt^2
128=16t^2
t^2=8
t<3 because it's more than 2^(1/2) but less than 3^(1/); much closer to 3