P11SnowDayNotes.notebook
February 06, 2017
To work with d=vit + 0.5at2 you need the value of t. In some problems, that isn't given to you but instead you have vf . You can find the time using t= (vf ­ vi)/a but if the problem doesn't require that as part of the answer, you can take a shortcut by creating a general formula that eliminates the need for t. Substitute the expression for t from the acceleration definition, into the equation for distance:
d=vit + 0.5at2 and t= (vf ­ vi)/a gives d=vi(vf ­ vi)/a + 0.5a((vf ­ vi)/a)2 which looks like a huge mess, until we use algebra to simplify it. Start by clearing some brackets by multiplying, using distribution , then cancel an 'a' factor in the second term :
multiply both sides by 'a' and by 2
to get rid of the denom. and 0.5
collect like terms
2ad = vf2 ­ vi2 a simpler formula with no time !
Isolating each term to get the different forms:
d=(vf2 ­ vi2)/(2a) or a=(vf2 ­ vi2)/(2d) Note : For these be sure to use the brackets in both numerator and denominator when entering into calculator.
vf2 = vi2 + 2ad or vi2 = vf2 ­ 2ad Note : For these the answer will still need to be square rooted to get the velocity.
There is one other thing to be aware of : you need to keep signs on your values that make sense, and in the case of the last two forms where you will be square­rooting , A) you cannot take the square root of a negative value, and B) your answer might be the negative root, not the positive one (remember if x2 = 4 then x=+2 or ­2)
As an example: gravity is downwards, giving a=­9.8 m/s2 and if I throw a ball upwards at 25 m/s I will define up as positive. As the ball slows down, it goes higher. What height is it at when it has slowed to 20 m/s? In other words, find the distance it has travelled if vf= 20, vi = 25 and a=­9.8 .
d= (202 ­ 252) / (2 * ­9.8) notice right away we will have a negative value in both numer. and denom.
d= (­225 ) / ( ­19.6) = + 11.479... or 11.5 m high, a positive answer as expected. How high will it go before it stops? (What is the maximum height it reaches before starting to fall back?)
"Stops" means a velocity of zero, d = (02 ­ 252)/(2*­9.8) = 31.887...m is the maximum height.
Now it starts to fall and picks up velocity downwards, now a negative value. How high is it when it hits a velocity of ­20 m/s ? Same formula, d= ((­20)2 ­ 252) / (2 * ­9.8) and you notice when we square the negative 20 we still get positive 400, then subtract 625 and get the same answer as before ­ the ball is again 11.5 m above us, now coming down at 20 m/s. What velocity will it have as it returns to its original height in our hand? Its height is now zero, vf2 = vi2 + 2ad where d = 0 so vf2 = vi2 + 0 and if you square root both sides vf = vi , right?
Not quite, since vi was up and positive, and we need the negative square root to get it right, the answer is it returns to our hand going ­25 m/s (same speed opposite direction). Now let the ball fall an extra 1.25 m to the ground. What is its velocity now? Logically we know it must be faster than the 25 m/s since it has had a little extra time accelerating, and we know it is still negative since it is going down. How to make sure the math works out? Again, take the negative root at the last step, but also put d=­1.25 (emphasis NEGATIVE) since this is 1.25 BELOW our original height. vf2 = vi2 + 2ad vf2 = (20) 2 + 2(­9.8) (­1.25) vf2 = 424.5 and v= ­20.60...m/s 1
P11SnowDayNotes.notebook
February 06, 2017
More Questions for Homework
13) Dropping a ball from the ceiling in class, what speed does it have when it has fallen 1.5m ? (recall dropping means vi=0)
14) Dropping a ball from the ceiling in class, what speed does it have when it has fallen 3.0m ?
15) Dropping a ball out the window in class, what speed does it have when it has fallen 6.0m ?
16) A jet on takeoff accelerates from 2.0 m/s to 40.0 m/s at a rate of 6.5 m/s2 ­ how much distance does it cover?
17) If a car can decelerate at 12.5 m/s2 maximum, what distance is required to stop from a speed of 11 m/s?
18) If a car can decelerate at 12.5 m/s2 maximum, what distance is required to stop from a speed of 22 m/s?
19) What is the maximum height of a ball thrown straight up with an initial speed of 8.57 m/s?
20) A car accelerates at 5.5 m/s2 for a distance of 45 m. If it started from a speed of 6.3 m/s, what is its final velocity? 21) A sprinter goes from 0 m/s to 12 m/s in the first 25 m of a race. What is the acceleration?
22) Astronomers observe a meteor approaching Jupiter, speeding up from 185 m/s to 387 m/s over a distance of 4.6 km. What is the acceleration? 23) What initial speed is required to launch a ball straight up a maximum distance of exactly 2.90 m ? 24) Skid marks at an accident site indicate a car decelerated with a=­11.5 m/s2 If the skid marks measure 42 m long what was the car's initial speed before skidding?
25) A ball thrown straight up at 12.5 m/s near the edge of the roof of a tall building is not caught on its way down, and continues until it hits the sidewalk 85 m below the height at which it was launched. a) What is the speed just at the moment of impact? b) What was the maximum height reached?
c) If it had just fallen from that maximum height instead (all the way to the sidewalk) how long would it have taken? d) Using vi = 12.5 and vf = your answer from part a), and a=­9.8 m/s2 what is the time needed for that acceleration? e) Explain the reason for the difference between answers c) and d), which should be about 1.2755... seconds
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