Question

If a rock is thrown upward on the planet Mars with a velocity of , its height in meters t seconds later is given by y=10t-1.86t^2 . Find the average velocity over the given time intervals:

Answer 1

(a). The average speed of the particle in the interval
`[1,2]` is
`\boxed{4.42\text{ m/s}}`.

(b). The average speed of the particle in the interval
`[1,1.5]` is
`\boxed{5.35\text{ m/s}}`.

Further Explanation:

The position of the particle on the surface of mars is given by:

`y=10t-1.86t^2`.

The average speed of the particle is given as:

`v_(avg)=\frac{\text{distance covered in the interval}}{\text{time taken in the inetrval}}`

Part (a):

The average speed of the particle in the interval
`\bf[1,2]`,

`v_(avg)=(y(2)-y(1))/(2-1)`

Now,

`\begin{aligned}y(2)&=10(2)-1.86(2^2)\\&=12.56\end{aligned}`

`\begin{aligned}y(1)&=10(1)-1.86(1^2)\\&=8.14\end{aligned}`

Substitute the values in above expression of
`v_(avg)`:

`\begin{aligned}v_(avg)&=(12.56-8.14)/(2-1)\\&=(4.42)/(1)\\&=4.42\text{ m/s}\end{aligned}`

So, the average speed of the particle in the interval
`[1,2]` is
`\boxed{4.42\text{ m/s}}`.

Part (b):

The average speed of the particle in the interval
`\bf[1,1.5]`,

`v_(avg)=(y(1.5)-y(1))/(1.5-1)`

Now,

`\begin{aligned}y(1.5)&=10(1.5)-1.86(1.5^2)\\&=10.815\end{aligned}`

`\begin{aligned}y(1)&=10(1)-1.86(1^2)\\&=8.14\end{aligned}`

Substitute the values in above expression of
`v_(avg)`:

`\begin{aligned}v_(avg)&=(10.815-8.14)/(1.5-1)\\&=(2.675)/(0.5)\\&=5.35\text{ m/s}\end{aligned}`

So, the average speed of the particle in the interval
`[1,1.5]` is
`\boxed{5.35\text{ m/s}}`.