Question

Here so I will need you to do a quick question for tomorrow thank

Answer 1

`\begin{gathered} y=58t-1.86t^2 \\ where\text{ y=height \lparen in meters\rparen} \\ and\text{ t = time in seconds} \end{gathered}`
`\begin{gathered} We\text{ want to find the average speed over the interval \lbrack1, 1.5\rbrack} \\ Height\text{ when t = 1 = 58\lparen1\rparen-1.86\lparen1}^2) \\ y=58-1.86 \\ y=56.14m \end{gathered}`
`\begin{gathered} Height\text{ when t = 1.5:} \\ y=58(1.5)\text{ -1.86 \lparen1.5\rparen}^2 \\ y=87-4.186 \\ y=82.814m \end{gathered}`
`\begin{gathered} Average\text{ speed = }\frac{total\text{ distance}}{total\text{ time taken}} \\ Average\text{ speed = }(56.14+82.814)/(1+1.5)\text{ = }(138.954)/(2.5) \\ =55.5816m\text{/s} \\ =55.58m\text{/s \lparen2 decimal places\rparen} \end{gathered}`

The answer is 55.58m/s