Physics help, acceleration question

[BeMD13]

Advertisement - Members don't see this ad

Can someone help me figure out how the steps to the correct answer? The question is...

An object that is initially moving in a straight line towards the left with a speed of 10 m/s experiences a constant acceleration for 4 s and ends up moving to the right with a speed of 2m/s. During that 4 s interval, what was the acceleration of the object?

The correct answer is supposed to be 3 m/s^2 to the right. I am not sure what it is but I am just completely stumped trying to get there. Is the second part with the 2m/s included in the first 4 seconds?

 

[sps27]

Can someone help me figure out how the steps to the correct answer? The question is...

An object that is initially moving in a straight line towards the left with a speed of 10 m/s experiences a constant acceleration for 4 s and ends up moving to the right with a speed of 2m/s. During that 4 s interval, what was the acceleration of the object?

The correct answer is supposed to be 3 m/s^2 to the right. I am not sure what it is but I am just completely stumped trying to get there. Is the second part with the 2m/s included in the first 4 seconds?

v = u + at v=final velocity, u=initial velocity
2 = -10 + 4a
a = 3m/s^2 (+ve) means to the right.

The basic idea is direction. Take to the left as -ve and to the right as +ve. I think your problem is more conceptual than mathematical. So the way I understand this is body is moving to the left and then a force acts on the body in exactly the opposite direction of movement decelerating it to come to stop and the force continues to act so it changes direction and start moving in the opposite direction. Now they only tell you the end conditions, i.e., velocity and total time, so I would just treat it like that making sure that I have the signs correct. I am not sure if this even helps you.....but this is what I think.....

 

[smiley27]

Can someone help me figure out how the steps to the correct answer? The question is...

An object that is initially moving in a straight line towards the left with a speed of 10 m/s experiences a constant acceleration for 4 s and ends up moving to the right with a speed of 2m/s. During that 4 s interval, what was the acceleration of the object?

The correct answer is supposed to be 3 m/s^2 to the right. I am not sure what it is but I am just completely stumped trying to get there. Is the second part with the 2m/s included in the first 4 seconds?

Hi BeMD13,

A simpler way of thinking about these types of constant acceleration problems is this:

Acceleration is equal to the change in velocity divided by the change in time. Or a is equal to delta v or delta t. You already know what delta t is, it's 4 seconds. But what is delta v? Delta v is vf - vi. What is vf and vi? How about you arbitrarily set the positive direction as being to the left because that's where the object is initially going at first.
Thus, vi is equal to +10m/s. What is vf? Since the object is going in the opposite direction at the end, we set vf equal to -2m/s.

Now we calculate accelerate. a is equal to (vf-vi)/t or (-2 - 10)/4 or -12/4 which is equal to -3m/s^2. Since the object changed its velocity, going initially to the left then finally to the right, we know that acceleration must be negative (or "against") the initial velocity because it eventually slows down and changes direction.

Another angle to take in helping us to understand the problem in a little more depth is this:

Assuming we already know the answer that acceleration is equal to -3 meters per second squared, or -3 meters per second, per second. Try to understand what that really means. -3 meters per second, PER SECOND. This means that, for ever second, the velocity is changing at -3 meters per second. Or simply: every second the velocity is changing by 3.

Thus, initially we have the velocity is equal to +10m/s, after the first second the velocity is +7m/s, after the 2nd second the velocity is +4m/s, after the third second the velocity is +1m/s, and finally after the fourth second the final velocity is -2m/s.

 

[BeMD13]

Oh wow, ok... after seeing solutions typed out, I am feeling like such a *****. I was not able to visualize this at all and I was making it much harder than it really was! Thanks for the help!