Question

If an arrow is shot upward on Mars with a speed of 62 m/s, its height in meters t seconds later is given by y = 62t − 1.86t². (Round your answers to two decimal places.) Estimate the speed when t = 1. Can you please show me the steps to solve this?

Answer 1

58.28 m/s

Explanation:

y = 62t - 1.86t²

Speed, S = dy/dt = 62 - 2(1.86)t

S = 62 - 3.72t

When t = 1

S = 62 - 3.72 = 58.28 m/s

Answer 2

Approximately
`58.28\; \rm m \cdot s^(-1)`.

Explanation:

The velocity of an object is the rate at which its position changes. In other words, the velocity of an object is equal to the first derivative of its position, with respect to time.

Note that the arrow here is launched upwards. (Assume that the effect of wind on Mars is negligible.) There would be motion in the horizontal direction. The horizontal position of this arrow will stays the same. On the other hand, the vertical position of this arrow is the same as its height:
`y = 62\, t - 1.86\, t^2`.

Apply the power rule to find the first derivative of this
`y` with respect to time
`t`.

By the power rule:

• the first derivative of
  `t` (same as
• the first derivative of
  `t^2` (same as
  `t` to the second power) with respect to

Therefore:

`\begin{aligned}(dy)/(d t) &= (d)/(d t)\left[62 \, t - 1.86\, t^2\right] \\ &= 62\,\left((d)/(d t)\left[t\right]\right) - 1.86\, \left((d)/(d t)\left[t^2\right]\right) \\ &= 62 * 1 - 1.86*\left(2\, t) = 62 - 3.72\, t\end{aligned}`.

In other words, the (vertical) velocity of this arrow at time
`t` would be
`(62 - 3.72\, t)` meters per second.

Evaluate this expression for
`t = 1` to find the (vertical) velocity of this arrow at that moment:
`62 - 3.72 * 1 =58.28`.