Question

If a rock is thrown upward on the planet Mars with a velocity of \(10{m}/{s}\;\),its height in meters tseconds later is given by\(y = 10t - 1.86{t^2}\).

(a) Find the average velocity over the given time intervals:

(i) (1, 2)

(ii) (1, 1.5)

(iii) (1, 1.1)

(iv) (1, 1.01)

(v) (1, 1.001)

(b) Estimate the instantaneous velocity when t = 1.

Short answer

(a) The average velocity for the given time interval is given below:

(i) \(4.42{\text{m}}/{\text{s}}\;\)

(ii) \(\text{5}\text{.35}{\text{m}}/{\text{s }}\;\)

(iii) \(6.094{\text{m}}/{\text{s}}\;\)

(iv) \(6.2614{\text{m}}/{\text{s}}\;\)

(v) \(6.2781{\text{m}}/{\text{s}}\;\)

(b) The instantaneous velocity at t = 1is \(6.28{\text{m}}/{\text{s}}\;\).

Step-by-step solution

Introduction

The average velocity of any object or material,

\({\rm{Avg}}{\rm{. vel}} = \frac{{{y_2} - {y_1}}}{{{t_2} - {t_1}}}\)

The instantaneous velocity of the object or material,

\({\rm{Ins}}{\rm{. vel}} = \frac{{dy}}{{dt}}\)

Given

Height \(y = 10t - 1.86{t^2}\)

Beginning time t = 1

Velocity \(v=10{m}/{s}\;\)

(a) (i)At time interval (1, 2) ,the average velocity.

Here,\({t_1} = 1,{t_2} = 2\)

\(\begin{align} & \text{Avg}\text{. vel}=\frac{{{y}_{2}}-{{y}_{1}}}{{{t}_{2}}-{{t}_{1}}}\\& =\frac{\left( 10\times 2-1.86\times {{2}^{2}} \right)-\left( 10\times 1-1.86\times {{1}^{2}} \right)}{1}\\& =4.42{\text{m}}/{\text{s}}\; \end{align}\)

(a) (ii)At time interval (1, 1.5) ,the average velocity.

Here,\({t_1} = 1,{t_2} = 1.5\)

\(\begin{align} & \text{Avg}\text{. vel}=\frac{{{y}_{2}}-{{y}_{1}}}{{{t}_{2}}-{{t}_{1}}} \\ & =\frac{\left( 10\times 1.5-1.86\times {{1.5}^{2}} \right)-\left( 10\times 1-1.86\times {{1}^{2}} \right)}{0.5} \\ & =5.35{\text{m}}/{\text{s}}\; \end{align}\)

(a) (iii)At time interval (1, 1.1) ,the average velocity.

Here,\({t_1} = 1,{t_2} = 1.1\)

\(\begin{align} & \text{Avg}\text{. vel}=\frac{{{y}_{2}}-{{y}_{1}}}{{{t}_{2}}-{{t}_{1}}} \\ & =\frac{\left( 10\times 1.1-1.86\times {{1.1}^{2}} \right)-\left( 10\times 1-1.86\times {{1}^{2}} \right)}{0.1} \\ & =6.094{\text{m}}/{\text{s}}\; \end{align}\) \(\)

(a) (iv)At time interval   (1, 1.01) ,the average velocity.

Here,\({t_1} = 1,{t_2} = 1.01\)

\(\begin{align} & \text{Avg}\text{. vel}=\frac{{{y}_{2}}-{{y}_{1}}}{{{t}_{2}}-{{t}_{1}}} \\ & =\frac{\left( 10\times 1.01-1.86\times {{1.01}^{2}} \right)-\left( 10\times 1-1.86\times {{1}^{2}} \right)}{0.01} \\ & =6.2614{\text{m}}/{\text{s}}\; \end{align}\)

(a)(v)At time interval   (1, 1.001) ,the average velocity.

Here,\({t_1} = 1,{t_2} = 1.001\)

\(\begin{align} & \text{Avg}\text{. vel}=\frac{{{y}_{2}}-{{y}_{1}}}{{{t}_{2}}-{{t}_{1}}} \\ & =\frac{\left( 10\times 1.001-1.86\times {{1.001}^{2}} \right)-\left( 10\times 1-1.86\times {{1}^{2}} \right)}{0.001} \\ & =6.2781{\text{m}}/{\text{s}}\; \end{align}\).

The instantaneous velocity at t = 1.

\(\begin{align} & \text{Ins}\text{. vel}=\frac{d\left(10t-1.86{{t}^{2}} \right)}{dt} \\ & =10-3.72t \\ & \text{Ins}\text{. vel}=10-3.72\times 1 \\ & =6.28{\text{m}}/{\text{s}}\; \end{align}\)