Problem Solving with Velocity and Acceleration
Important Formulas:
= da = 2 - 1a = d = 1 + 2 t
t t t 2
= d av = 1 + 2 or total displacementd = 1t + ½ at2
t 2 total time
Slope = rise= y2 – y1A = ½ b x h (triangle)A = b x h (rectangle)
run x2 – x1
dr = d = d1 + d2 + …..
Working With Accelerations
Example 1:
A car starts from rest and accelerates at 2 m/s/s for 5 seconds. How fast will it be going?
Given:
The statement "car starts from rest" means that the car's starting velocity is 0 m/s. An acceleration of 2 m/s/s means that the car's velocity changes by 2 m/s each second.
Required: velocity
Analysis and Solution:
then , so.
Sentence:
If the velocity starts at 0 m/s and changes by 10 m/s, it ends at 10 m/s.
Working to find Displacement
Example 2:
Rennata Gas is driving through town at 25.0 m/s and begins to accelerate at a constant rate of -1.0 m/s2. Eventually Rennata comes to a complete stop.
a.Represent Rennata's accelerated motion by sketching a velocity-time graph. Use the velocity-time graph to determine this displacement.
b.Use kinematic equations to calculate the displacement which Rennata travels while decelerating.
Solution:
a)The velocity-time graph for the motion is:
The displacement traveled can be found by a calculation of the area between the line on the graph and the time axis.
Area = 0.5*b*h = 0.5*(25.0 s)*(25.0 m/s)
Area = 313 m
b)The displacement traveled can be calculated using a kinematic equation. The solution is shown here.
Given:vi = 25.0 m/s / vf = 0.0 m/s / a = -1.0 m/s2
/ Find:
d = ??
Analysis and Solution:
vf2 = vi2 + 2*a*d
(0 m/s)2 = (25.0 m/s)2 + 2 * (-1.0 m/s2)*d
0.0 m2/s2 = 625.0 m2/s2 + (-2.0 m/s2)*d
0.0 m2/s2 - 625.0 m2/s2 = (-2.0 m/s2)*d
(-625.0 m2/s2)/(-2.0 m/s2) =d
313 m =d
Sentence:
Therefore the displacement is 313 m.
Example 2:
Otto Emissions is driving his car at 25.0 m/s. Otto accelerates at 2.0 m/s2 for 5 seconds. Otto then maintains a constant velocity for 10.0 more seconds.
a.Represent the 15 seconds of Otto Emission's motion by sketching a velocity-time graph. Use the graph to determine the displacement which Otto traveled during the entire 15 seconds.
b.Finally, break the motion into its two segments and use kinematic equations to calculate the total displacement traveled during the entire 15 seconds.
Solution to Question 2
a.The velocity-time graph for the motion is:
The displacement traveled can be found by a calculation of the area between the line on the graph and the time axis. This area would be the area of the triangle plus the area of rectangle 1 plus the area of rectangle 2.
Area = 0.5*b1*ht + b1*h1 + b2*h2
Area = 0.5*(5.0 s)*(10.0 m/s) + (5.0 s)*(25.0 m/s) + (10.0 s)*(35.0 m/s)
Area = 25 m + 125 m + 350 m
Area = 500 m
b.The displacement traveled can be calculated using a kinematic equation. The solution is shown here.
First find the d for the first 5 seconds:
Given:vi = 25.0 m/s / t = 5.0 s / a = 2.0 m/s2
/ Find:
d = ??
d = vi*t + 0.5*a*t2
d = (25.0 m/s)*(5.0 s) + 0.5*(2.0 m/s2)*(5.0 s)2
d = 125 m + 25.0 md = 150 m
Now find the d for the last 10 seconds:
Given:vi = 35.0 m/s / t = 10.0 s / a = 0.0 m/s2
/ Find:
d = ??
(Note: the velocity at the 5 second mark can be found from knowing that the car accelerates from 25.0 m/s at +2.0 m/s2 for 5 seconds. This results in a velocity change of a*t = 10 m/s, and thus a velocity of 35.0 m/s.)
d = vi*t + 0.5*a*t2
d = (35.0 m/s)*(10.0 s) + 0.5*(0.0 m/s2)*(10.0 s)2
d = 350 m + 0 m d =350 m
The total displacement for the 15 seconds of motion is the sum of these two displacement calculations (150 m + 350 m): displacement = 500 m
Therefore the displacement is 500 m.
Remember, generally speaking most problem solving situations in this course will involve a constant acceleration.