An Alternate Solution for Example 2


[Up][Home][Help][Chapter 2 Objectives]

BHS -> Staff -> Mr. Stanbrough -> Physics -> Mechanics -> Kinematics -> Using v = vo + at -> this page


Example 2:

A boat, initially moving at 20 m/s is only moving at 2.0 m/s in the same direction, 25 seconds later. Assuming that the boat's acceleration was constant, what was the boat's acceleration during this time?

Solution:

Diagram for example 2

vo = 20 m/s

v = 2.0 m/s

delta t = 25 s

a = ?

Here's the plan: you can find the acceleration from the equation a = delta_v/delta_t, but we don't know delta v. However, you can find delta v from the equation delta v = v - vo, then substitute...

a = delta_v/delta_t, but:

delta_v = v - vo

delta v = 2.0 m/s - 20 m/s = -18 m/s

solution

The boat's acceleration is 0.72 m/s2, in the direction opposite to the boat's velocity.

Note:

This is a 2-step solution since you have to find delta v as an intermediate step, but each step is pretty simple.


[Up][Home][Help][Chapter 2 Objectives]

BHS -> Staff -> Mr. Stanbrough -> Physics -> Mechanics -> Kinematics -> Using v = vo + at -> this page


last update August 1, 2001 by JL Stanbrough