A particle has an initial horizontal velocity of 1.9 m/s and an initial upward velocity of 3.7 m/s. It is then given a horizonta
l acceleration of 1.1 m/s 2 and a downward acceleration of 1.2 m/s 2 . What is its speed after 3.4 s? Answer in units of m/s.
1 answer:
Its horizontal and vertical velocities are given respectively by
![v_x=1.9\dfrac{\rm m}{\rm s}+\left(1.1\dfrac{\rm m}{\mathrm s^2}\right)t](https://tex.z-dn.net/?f=v_x%3D1.9%5Cdfrac%7B%5Crm%20m%7D%7B%5Crm%20s%7D%2B%5Cleft%281.1%5Cdfrac%7B%5Crm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t)
![v_y=3.7\dfrac{\rm m}{\rm s}+\left(-1.2\dfrac{\rm m}{\mathrm s^2}\right)t](https://tex.z-dn.net/?f=v_y%3D3.7%5Cdfrac%7B%5Crm%20m%7D%7B%5Crm%20s%7D%2B%5Cleft%28-1.2%5Cdfrac%7B%5Crm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t)
After
, the components of its velocity are
![v_x=1.9\dfrac{\rm m}{\rm s}+\left(1.1\dfrac{\rm m}{\mathrm s^2}\right)(3.4\,\mathrm s)=5.64\dfrac{\rm m}{\rm s}](https://tex.z-dn.net/?f=v_x%3D1.9%5Cdfrac%7B%5Crm%20m%7D%7B%5Crm%20s%7D%2B%5Cleft%281.1%5Cdfrac%7B%5Crm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29%283.4%5C%2C%5Cmathrm%20s%29%3D5.64%5Cdfrac%7B%5Crm%20m%7D%7B%5Crm%20s%7D)
![v_y=3.7\dfrac{\rm m}{\rm s}+\left(-1.2\dfrac{\rm m}{\mathrm s^2}\right)(3.4\,\mathrm s)=-0.38\dfrac{\rm m}{\rm s}](https://tex.z-dn.net/?f=v_y%3D3.7%5Cdfrac%7B%5Crm%20m%7D%7B%5Crm%20s%7D%2B%5Cleft%28-1.2%5Cdfrac%7B%5Crm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29%283.4%5C%2C%5Cmathrm%20s%29%3D-0.38%5Cdfrac%7B%5Crm%20m%7D%7B%5Crm%20s%7D)
Its overall speed is the magnitude of its velocity:
![\|\vec v\|=\sqrt{{v_x}^2+{v_y}^2}\approx5.7\dfrac{\rm m}{\rm s}](https://tex.z-dn.net/?f=%5C%7C%5Cvec%20v%5C%7C%3D%5Csqrt%7B%7Bv_x%7D%5E2%2B%7Bv_y%7D%5E2%7D%5Capprox5.7%5Cdfrac%7B%5Crm%20m%7D%7B%5Crm%20s%7D)
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