Respuesta :
(a). The average speed of the particle in the interval [tex][1,2][/tex] is [tex]\boxed{4.42\text{ m/s}}[/tex].
(b). The average speed of the particle in the interval [tex][1,1.5][/tex] is [tex]\boxed{5.35\text{ m/s}}[/tex].
Further Explanation:
The position of the particle on the surface of mars is given by:
[tex]y=10t-1.86t^2[/tex].
The average speed of the particle is given as:
[tex]v_{avg}=\dfrac{\text{distance covered in the interval}}{\text{time taken in the inetrval}}[/tex]
Part (a):
The average speed of the particle in the interval [tex]\bf[1,2][/tex],
[tex]v_{avg}=\dfrac{y(2)-y(1)}{2-1}[/tex]
Now,
[tex]\begin{aligned}y(2)&=10(2)-1.86(2^2)\\&=12.56\end{aligned}[/tex]
[tex]\begin{aligned}y(1)&=10(1)-1.86(1^2)\\&=8.14\end{aligned}[/tex]
Substitute the values in above expression of [tex]v_{avg}[/tex]:
[tex]\begin{aligned}v_{avg}&=\dfrac{12.56-8.14}{2-1}\\&=\dfrac{4.42}{1}\\&=4.42\text{ m/s}\end{aligned}[/tex]
So, the average speed of the particle in the interval [tex][1,2][/tex] is [tex]\boxed{4.42\text{ m/s}}[/tex].
Part (b):
The average speed of the particle in the interval [tex]\bf[1,1.5][/tex],
[tex]v_{avg}=\dfrac{y(1.5)-y(1)}{1.5-1}[/tex]
Now,
[tex]\begin{aligned}y(1.5)&=10(1.5)-1.86(1.5^2)\\&=10.815\end{aligned}[/tex]
[tex]\begin{aligned}y(1)&=10(1)-1.86(1^2)\\&=8.14\end{aligned}[/tex]
Substitute the values in above expression of [tex]v_{avg}[/tex]:
[tex]\begin{aligned}v_{avg}&=\dfrac{10.815-8.14}{1.5-1}\\&=\dfrac{2.675}{0.5}\\&=5.35\text{ m/s}\end{aligned}[/tex]
So, the average speed of the particle in the interval [tex][1,1.5][/tex] is [tex]\boxed{5.35\text{ m/s}}[/tex].
