1.A hiker started to walk from her campsite heading east with a velocity of 5 m\s after 6 seconds, she attained a velocity of 7 m\s. What is the average velocity of the hiker? What is the hiker’s acceleration? If the hiker continues walking for another 60 seconds, how far east can she go?

a. avgvelocity=(final+starting)/2= 6m/s

acceleration=changevelocity/time=1/3 m/s^2

distance=distance first six seconds+distance for last 54 seconds
= avgvelocity*6+7*54=36m+378m

456

To find the average velocity of the hiker, you need to divide the total displacement by the total time taken.

The initial velocity (u) of the hiker is 5 m/s, and after 6 seconds, the final velocity (v) is 7 m/s.

The average velocity (𝑣̅) can be calculated using the formula:

𝑣̅ = (𝑢 + 𝑣)/2

Substituting the given values:

𝑣̅ = (5 + 7)/2
= 12/2
= 6 m/s

So, the average velocity of the hiker is 6 m/s.

Now, to find the acceleration (a) of the hiker, you can use the formula:

a = (v - u)/t

Substituting the given values:

a = (7 - 5)/6
= 2/6
= 1/3 m/s²

The hiker's acceleration is 1/3 m/s².

To calculate the distance the hiker can go in the next 60 seconds, you need to use the equation of motion:

𝑠 = 𝑢𝑡 + 1/2𝑎𝑡²

In this case, since the hiker's acceleration is not given, we can assume it to be constant and equal to the calculated value above (1/3 m/s²).

Substituting the values:

𝑠 = 5 × 60 + 1/2 × 1/3 × (60)²
= 300 + 1/2 × 1/3 × 3600
= 300 + 1/6 × 3600
= 300 + 600
= 900 meters

Therefore, the hiker can travel 900 meters east in the next 60 seconds.