Chapter 2.3, Problem 15PP

(a)

To determine

Time at which A was at a distance of $\text{25 m}$ from cafeteria.

Answer to Problem 15PP

  $t\text{=}19.23\text{ s}$

Explanation of Solution

Given:

Distance between cafeteria and current position, $x=\text{25 m}$ .

Distance traveled in $2\text{ s}$ is $\text{2}\text{.6 m}$ .

Formula used:

  $\text{Speed=}\frac{\text{distance}}{\text{time}}$

Calculation:

Distance traveled in $2\text{ s}$ is $\text{2}\text{.6 m}$ .

  $\text{Speed=}\frac{\text{2}\text{.6 m}}{2\text{ s}}=1.3\text{ m/s}$

Time taken to cover $\text{25 m}$

  $\text{time=}\frac{\text{distance}}{\text{speed}}$

  $t=\frac{25\text{ m}}{1.3\text{ m/s}}$

  $t\text{=}19.23\text{ s}$

Conclusion:

A travelled a distance of $\text{25 m}$ in $t\text{=}19.23\text{ s}$ .

(b)

To determine

Time at which A was at a distance of $\text{25 m}$ from Band room.

Answer to Problem 15PP

Time at which A was at a distance of $\text{25 m}$ from Band room is $t\text{=}57.7\text{ s}$ .

Explanation of Solution

Given:

Distance between Band room and current position, $x=\text{25 m}$

Distance traveled in $2\text{ s}$ is $\text{2}\text{.6 m}$ .

Formula used:

  $\text{Speed=}\frac{\text{distance}}{\text{time}}$

Calculation:

  $\text{Speed=}\frac{\text{distance}}{\text{time}}$

Distance traveled in $2\text{ s}$ is $\text{2}\text{.6 m}$ .

  $\text{Speed=}\frac{\text{2}\text{.6 m}}{2\text{ s}}=1.3\text{ m/s}$

Distance covered= $\text{100 m-25 m=75 m}$

Time taken to cover $\text{75 m}$

  $\text{time=}\frac{\text{distance}}{\text{speed}}$

  $t=\frac{75\text{ m}}{1.3\text{ m/s}}$

  $t\text{=}57.7\text{ s}$

Conclusion:

Time at which A was at a distance of $\text{25 m}$ from Band room is $t\text{=}57.7\text{ s}$ .

(c)

To determine

Graph between A’s position and time.

Explanation of Solution

Given:

Distance moved is 2.6 m at every 2 s.

Calculation:

X (time in second)	Y(position in metres)
0	0
10	13
20	26
30	39
40	52
50	65
60	78
70	91
80	104
90	117
100	130

Graph:

Graph between Ari’s position and time is given as: