The acceleration of the particle is given by the formula mentioned below:
Differentiate the position vector with respect to t.
![\begin{gathered} \frac{ds(t)}{dt}=\frac{d}{dt}\sqrt[]{\mleft(t^3+1\mright)} \\ =-\frac{1}{2}(t^3+1)^{-\frac{1}{2}}\times3t^2 \\ =\frac{3}{2}\frac{t^2}{\sqrt{(t^3+1)}} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Cfrac%7Bds%28t%29%7D%7Bdt%7D%3D%5Cfrac%7Bd%7D%7Bdt%7D%5Csqrt%5B%5D%7B%5Cmleft%28t%5E3%2B1%5Cmright%29%7D%20%5C%5C%20%3D-%5Cfrac%7B1%7D%7B2%7D%28t%5E3%2B1%29%5E%7B-%5Cfrac%7B1%7D%7B2%7D%7D%5Ctimes3t%5E2%20%5C%5C%20%3D%5Cfrac%7B3%7D%7B2%7D%5Cfrac%7Bt%5E2%7D%7B%5Csqrt%7B%28t%5E3%2B1%29%7D%7D%20%5Cend%7Bgathered%7D)
Differentiate both sides of the obtained equation with respect to t.
![\begin{gathered} \frac{d^2s(t)}{dx^2}=\frac{3}{2}(\frac{2t}{\sqrt[]{(t^3+1)}}+t^2(-\frac{3}{2})\times\frac{1}{(t^3+1)^{\frac{3}{2}}}) \\ =\frac{3t}{\sqrt[]{(t^3+1)}}-\frac{9}{4}\frac{t^2}{(t^3+1)^{\frac{3}{2}}} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Cfrac%7Bd%5E2s%28t%29%7D%7Bdx%5E2%7D%3D%5Cfrac%7B3%7D%7B2%7D%28%5Cfrac%7B2t%7D%7B%5Csqrt%5B%5D%7B%28t%5E3%2B1%29%7D%7D%2Bt%5E2%28-%5Cfrac%7B3%7D%7B2%7D%29%5Ctimes%5Cfrac%7B1%7D%7B%28t%5E3%2B1%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%29%20%5C%5C%20%3D%5Cfrac%7B3t%7D%7B%5Csqrt%5B%5D%7B%28t%5E3%2B1%29%7D%7D-%5Cfrac%7B9%7D%7B4%7D%5Cfrac%7Bt%5E2%7D%7B%28t%5E3%2B1%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%20%5Cend%7Bgathered%7D)
Substitute t=2 in the above equation to obtain the acceleration of the particle at 2 seconds.
![\begin{gathered} a(t=1)=\frac{3}{\sqrt[]{2}}-\frac{9}{4\times2^{\frac{3}{2}}} \\ =1.32ft/sec^2 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20a%28t%3D1%29%3D%5Cfrac%7B3%7D%7B%5Csqrt%5B%5D%7B2%7D%7D-%5Cfrac%7B9%7D%7B4%5Ctimes2%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%20%5C%5C%20%3D1.32ft%2Fsec%5E2%20%5Cend%7Bgathered%7D)
The initial position is obtained at t=0. Substitute t=0 in the given position function.
10% of 50kg is 5kg
50kg add 5kg is 55kg
=55kg
Answer:
1) 20% change I think its an increase.
If needed I could try to answer the 2nd question but I am sure that I got the first one right.
Step-by-step explanation:
Answer:
b and c are right
Step-by-step explanation:
if its on odyssey ware that's the right ones
The mean time it takes to walk to the bus stop is 8 minutes (with a standard deviation of 2 minutes) and the mean time it takes for the bus to get to school is 20 minutes (with a standard deviation of 4 minutes). The distributions are normal.
a. How long will it take (in minutes), on average, to get to school?
b. What is the standard deviation of the trip to school?
c. What is the probability that it will take longer than 30 minutes to get to school?
Due to a miscalculation, you realize it actually takes an average of 10 minutes to walk to the bus stop.
d. How long will it take (in minutes), on average, to get to school?
e. What is the standard deviation of the trip to school?
f. What is the probability that it will take longer than 30 minutes to get to school?
The only ones I need help with is C and F. I have the answer for the rest of them. Can someone please help me with parts C and F??
