All you need to know is that velocity is the derivative with respect to time of position.
Therefore: ds/dt = -32t +10
To know the velocity at t=2s you just need to plug t=2 for t in the equation above.
Acceleration I think if I’m not mistaken
- Initial velocity (u) = 0 m/s [the car was at rest]
- Distance (s) = 80 m
- Time (t) = 10 s
- Let the magnitude of acceleration be a.
- By using the equation of motion,
we get,![80 = 0 \times 10 + \frac{1}{2} \times a \times {10}^{2} \\ = > 80 = \frac{1}{2} \times 100a \\ = > 80 = 50a \\ = > a = \frac{80}{50} \\ = > a = 1.6](https://tex.z-dn.net/?f=80%20%3D%200%20%5Ctimes%2010%20%2B%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%5Ctimes%20a%20%5Ctimes%20%20%7B10%7D%5E%7B2%7D%20%20%5C%5C%20%20%3D%20%20%3E%2080%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%5Ctimes%20100a%20%5C%5C%20%20%3D%20%20%3E%2080%20%3D%2050a%20%5C%5C%20%20%3D%20%20%3E%20a%20%3D%20%20%5Cfrac%7B80%7D%7B50%7D%20%20%5C%5C%20%20%3D%20%20%3E%20a%20%3D%201.6)
<u>A</u><u>nswer:</u>
<u>The </u><u>magnitude</u><u> </u><u>of </u><u>its </u><u>acceleration</u><u> </u><u>is </u><u>1</u><u>.</u><u>6</u><u> </u><u>m/</u><u>s^</u><u>2</u><u>.</u>
Hope you could get an idea from here.
Doubt clarification - use comment section.
They would weight 69N on the moon's surface.