#$&* course Mth 277 qa 10_03If the velocity function for a projectile is `v(t) = 10 `i + (20 - 9.8 t) `j, then:
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Given Solution: Velocity is the derivative of position, so you need an antiderivative. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q002. What is its position function if its t = 0 position is `R(0) = 0 `i + 10 `j? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `R(0) = 0 `i + 10 `j R(t)=(10t+c)i+(20t-4.9t^2+c)j R(0)=c1i+c2j c1=0 c2=10 R(t)=(10t)i+(20t-4.9t^2+10)j confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Your antiderivatives contain integration constants. From the given conditions you can evaluate those constants. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q003. At what instant is the `j component of the position function equal to 20? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: R(t)=(10t)i+(20t-4.9t^2+10)j j comp=20 20t-4.9t^2+10=20 -4.9t^2+20t-10=0 Quadr Form Used J comp is equal to 20 when t approximately equals .5834 and 4.532 units. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q004. At what instant is the `i component of the position equal to 20, and at that instant what is the `j component of its position? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: R(t)=(10t)i+(20t-4.9t^2+10)j 10t=20 t=2 units R(2)=20i+30.4j jcomp=30.4 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q005. At what instant is the `j component of its position maximized? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: R(t)=(10t)i+(20t-4.9t^2+10)j jcomp=20t-4.9t^2+10 deriv or v(t)=-9.8t+20=0 t is approx 2.041 units when jcomp os at its max which is approx 30.408 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: A function is maximized or minimized at a critical point. A first- or second-derivative test can check whether a critical point gives us a max or a min, or perhaps an inflection point. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q006. At what instant is the `j component of its position zero, and at that instant what is the `i component of its position? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: R(t)=(10t)i+(20t-4.9t^2+10)j Once again we need to use quad. -4.9^2+20t+10=0 one of the times approx equaled -.45 which does not make sense since the time cant be negative so we will discard that answer the other t was approx equal to 4.53 units which is reasonable and it checks out after plugging it into the original j comp for position. R(4.532)=45.32i+0j i comp=45.32 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The quadratic formula might be useful. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q007. At what instant is the angle between `R(t) and the `i vector equal to 70 degrees? Does this occur at only one instant? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: R(t) dot 'i= R(t)=(10t)i+(20t-4.9t^2+10)j 'i=(10t)i 'R(t) dot 'i=100t^2 cos(70)=100t^2/(sqrt((10t)^2+(20t)^2+(-4.9t^2)^2+100)sqrt((10t)^2)) cos(70)=100t^2/((49t^3+223.6t^2+100t)) .342=100t/(49t^2+223.6t+100) 16.758t^2+76.47t+34.2=100t 16.758t^2-25.53t+34.2=0 Quadr Form I must of made an error because I am getting a neg under the radical. I am thinking the a term should maybe be negative, but when I took the magnitude the a term would become positive.
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Given Solution: Use the dot product to get an expression for the angle. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q008. Give a set of parametric equations x = x(t) and y = y(t) that describe the position of the projectile. Eliminate the variable t, and solve for y in terms of x. What kind of equation do you get? Describe its graph. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: R(t)=(10t)i+(20t-4.9t^2+10)j t=x/10 Now I will plug into y y=-.49x^2+2x+10 As seen above, we get a quadratice equation and its shape is a parabola which opens downward. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The position is x(t) `i + y(t) `j. You figured out the position function in the second problem. You eliminate the variable by solving for either x or y in terms of t, then substituting in the equation for y or x (depending on whether you solve the x or the y equation for t). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!