Question

An astronaut has left the International Space Station to test a new space scooter. Her partner measures the following velocity changes, each taking place in a time interval 10.4s s . What are the average acceleration in each interval? Assume that the positive direction is to the right.
1.) At the beginning of the interval the astronaut is moving toward the right along the x-axis at 15.0 m/s, and at the end of the interval she is moving toward the right at 5.30 m/s.



a = ??



2.) At the beginning she is moving toward the left at 5.30 m/s. and at the end she is moving toward the left at 15.0 m/s.



a = ??



3.) At the beginning she is moving toward the right at 15.0 m/s, and at the end she is moving toward the left at 15.0 m/s.



a = ??



I have no clue how to do these problems. Please show step by step with correct answers so I can try to understand.

Answer 1

Acceleration can be defined as the change of speed in an instant of time, that is

$a = \frac{v_f-v_i}{\Delta t}$

Here,

$v_f =$ Final velocity

$v_i =$ Initial velocity

$\Delta t =$ Change in time

From this expression we will calculate the requested values replacing the variables in each of the given terms

PART A) The values under this condition are:

$v_f = 5.3m/s$

$v_i = 15m/s$

$\Delta t = 10.4s$

Replacing,

$a = \frac{5.3m/s-(15m/s)}{10.4s}$

$a = -0.93m/s^2$

PART B ) The values under this condition are:

$v_f = -15m/s$

$v_i = -5.3m/s$

$\Delta t = 10.4$

Replacing,

$a = \frac{-(15m/s)-(-5.3m/s)}{10.4s}$

$a = -0.93m/s^2$

Therefore the acceleration in the second time interval is $-0.93m/s^2$

PART C) The values under this condition are:

$v_f = -15m/s$

$v_i = 15m/s$

$\Delta t = 10.4$

Replacing,

$a = \frac{-15m/s-(15m/s)}{10.4s}$

$a = -2.9m/s^2$