Question

the scale on the horizontal axis is 3s per division and the verticle axis 8m/s per division.
with the same initial position of 23m, what is the acceleration when t=9s? answer in untis of m/s^2

Answer 1

Based off of your graph, the acceleration is zero m/s^2. If we consider the definition of an acceleration, then we know that an acceleration is an object speeding up, slowing down, or changing direction. The vertical scale represents 8 m/s per division. As a result our starting initial velocity would be 8*4 = 32 m/s. Because there is no increase or decrease in the velocity, the slope of the line is zero. Because the slope of a line is calculated by taking the change in variables in the y-axis and dividing those by the change in variables in the x-axis, our units are m/s divided by s which yields the unit of acceleration which is m/s^2. As a result the acceleration must be 0 m/s^2.

Does this make sense or would you like further explanation?

Answer 2

oh ok so what if they ask you for a position instead of the acceleration?

Answer 3

Well, you mean for a velocity vs time graph then asking you for position or a position vs time graph?

Answer 4

velocity vs time graph

Answer 5

Ok. So you can determine your initial velocity from your t=0s point on the graph correct? You can then determine your final velocity by seeing where the final time matches up with which final velocity value. So you would know your initial velocity, final velocity, and change in time. These are 3 of the 5 different quantities we need to know for kinematics (the other two being acceleration and displacement). To find the acceleration, we can use the equation:
a = (vf - vi)/t
which is actually just finding the slope of the line.
Then, choose your favorite kinematic equation to find the displacment. For example we could use the equation
\[\Delta x =v _{i} \Delta t + \frac{ 1 }{ 2 } a \Delta t ^{2}\]
or
\[\Delta x = \frac{ (v_{f}^{2} - v_{i}^{2}) }{ 2a }\]

Its really no different than solving a word problem. We just need to know where to start. What do you think?

Answer 6

oh ok for the acceleration, since the initial velocity and the final velocity are the same wouldnt that make it 0/9s ?

Answer 7

yep. You are correct so the acceleration is 0 m/s^2.

Answer 8

ok so then how would you solve for the displacement...i got confused right now

Answer 9

?

Answer 10

Oh sorry. I'm here. 1 sec.

Answer 11

To solve for the displacement we could use the equation\[\Delta x = v _{i} \Delta t + \frac{ 1 }{ 2 } a \Delta t\]
Because our acceleration is zero, we are left with
\[\Delta x = v _{i} \Delta t\]
Our initial velocity is 32 m/s and our change in time is 24s????
How far does that line go? Does it go to the 8 0r 9?

Answer 12

8

Answer 13

So if its 8, then 8*3 is 24 b/c 3 sec/div * 8 div = 24 s.
Now all we then need to do is take 32 m/s * 24 s and tell me what you end up with for your displacement.

Answer 14

768m

Answer 15

That would be it then. If you can decipher what you're given in your graphs you can always apply them to the kinematic equations.

Answer 16

oh ok thank you

Answer 17

Anytime. Good Luck.