<u>P</u><u>e</u><u>r</u><u>s</u><u>o</u><u>n</u><u>-</u><u>1</u>
- Initial velocity=u=0m/s
- Final velocity=v=10m/s
- Time=10s=t
![\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20Acceleration%3D%5Cdfrac%7Bv-u%7D%7Bt%7D)
![\\ \sf\longmapsto Acceleration=\dfrac{10-0}{10}](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20Acceleration%3D%5Cdfrac%7B10-0%7D%7B10%7D)
![\\ \sf\longmapsto Acceleration=\dfrac{10}{10}](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20Acceleration%3D%5Cdfrac%7B10%7D%7B10%7D)
![\\ \sf\longmapsto Acceleration=1m/s^2](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20Acceleration%3D1m%2Fs%5E2)
<u>P</u><u>e</u><u>r</u><u>s</u><u>o</u><u>n</u><u>-</u><u>2</u>
- initial velocity=0m/s=u
- Final velocity=v=0.25m/s
- Time=t=2s
![\\ \sf\longmapsto Acceleration=\dfrac{0.25-0}{2}](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20Acceleration%3D%5Cdfrac%7B0.25-0%7D%7B2%7D)
![\\ \sf\longmapsto Acceleration=\dfrac{0.25}{2}](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20Acceleration%3D%5Cdfrac%7B0.25%7D%7B2%7D)
![\\ \sf\longmapsto Acceleration=0.125m/s^2](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20Acceleration%3D0.125m%2Fs%5E2)
Person-1 is accelerating faster.
Answer:
![(b)\ t_1 - t_0](https://tex.z-dn.net/?f=%28b%29%5C%20t_1%20-%20t_0)
![(d)\ t_2 - t_1](https://tex.z-dn.net/?f=%28d%29%5C%20t_2%20-%20t_1)
![(e)\ \frac{t_2 - t_0}{2}](https://tex.z-dn.net/?f=%28e%29%5C%20%20%5Cfrac%7Bt_2%20-%20t_0%7D%7B2%7D)
Explanation:
Given
See attachment for complete question
Required
How long to reach the ground from the maximum height
First, calculate the time of flight (T)
![T =t_2 - t_0](https://tex.z-dn.net/?f=T%20%3Dt_2%20-%20t_0)
The time taken (t) from maximum height to the ground is:
![t = \frac{1}{2}T](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B1%7D%7B2%7DT)
So, we have:
![t = \frac{t_2 - t_0}{2}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7Bt_2%20-%20t_0%7D%7B2%7D)
Another representation is:
At ymax, the time is: t1
On the ground, the time is t2
The difference between these times is the time taken.
So;
![t = t_2 - t_1](https://tex.z-dn.net/?f=t%20%3D%20t_2%20-%20t_1)
Since air resistance is to be ignored, then
--- i.e. time to reach the maximum height from the ground equals time to reach the ground from the maximum height
Answer:
The second one is a function. {(-8, -2), (7, -2),(-9,2), (0,0)
Explanation:
Its because the y-value is repeated twice.
Hope it helps.
Answer:
Explanation:
m1 = 3.77 kg (0, 0 )
m2 = 6.7106 kg ( 5.72 cm, 11.44 cm)
m3 = 2.46181 kg (16.7024 cm, 0 cm )
Let x and y be the coordinates of centre of mass.
![x = \frac{m_{1}x_{1}+ m_{2}x_{2}+m_{3}x_{3}}{m_{1}+m_{2}+m_{3}}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7Bm_%7B1%7Dx_%7B1%7D%2B%20m_%7B2%7Dx_%7B2%7D%2Bm_%7B3%7Dx_%7B3%7D%7D%7Bm_%7B1%7D%2Bm_%7B2%7D%2Bm_%7B3%7D%7D)
![x = \frac{3.77\times 0+ 6.7106\times 5.72 + 2.46181\times 16.7024}{3.77+6.7106+2.46181}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B3.77%5Ctimes%200%2B%206.7106%5Ctimes%205.72%20%2B%202.46181%5Ctimes%2016.7024%7D%7B3.77%2B6.7106%2B2.46181%7D)
x = 6.1428 cm
![y = \frac{m_{1}y_{1}+ m_{2}y_{2}+m_{3}y_{3}}{m_{1}+m_{2}+m_{3}}](https://tex.z-dn.net/?f=y%20%3D%20%5Cfrac%7Bm_%7B1%7Dy_%7B1%7D%2B%20m_%7B2%7Dy_%7B2%7D%2Bm_%7B3%7Dy_%7B3%7D%7D%7Bm_%7B1%7D%2Bm_%7B2%7D%2Bm_%7B3%7D%7D)
![y= \frac{3.77\times 0+ 6.7106\times 11.44 + 2.46181\times 0}{3.77+6.7106+2.46181}](https://tex.z-dn.net/?f=y%3D%20%5Cfrac%7B3.77%5Ctimes%200%2B%206.7106%5Ctimes%2011.44%20%2B%202.46181%5Ctimes%200%7D%7B3.77%2B6.7106%2B2.46181%7D)
y = 5.9316 cm