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I'm currently working on a problem in which an object's speed remains the same, but it experiences a change in direction after some time. I'm fairly sure it experiences an acceleration in order to change direction.

However, I am confused because of the change in direction.

I want to use the formula $A_x = \Delta v / \Delta t $. Since the object experiences a change in direction at some point but still travels at the same speed, would the second velocity technically be negative? Like so:

$V_1$ = 5.0m/s ,$V_2$ = 5.0m/s, $T_1$ = 0 seconds, $T_2$ = 10 seconds

$A_x = V_1 - (-V_2) / T_1 - T_2 $

Also, does it matter which direction? Would it be the same as travelling right and then turning left/right/around?

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    $\begingroup$ My answer to Uniform circular motion may help you understand what is going on, though i don';t thionk it's a duplicate of your question. $\endgroup$ – John Rennie Jan 19 '15 at 16:50
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I'm fairly sure it experiences an acceleration in order to change direction.

Yes, any change in velocity means an acceleration has occurred. That includes a change in direction of the velocity.

Since the object experiences a change in direction at some point but still travels at the same speed, would the second velocity technically be negative?

If the first velocity is in the positive direction, yes the second might be negative. Is this a 1-dimensional problem?

I want to use the formula $A_x=\Delta v/ \Delta t.$

That will give you the average acceleration over the total time. It tells you nothing about the instantaneous acceleration at any specific time. Also, for $A_x$ you would use $\Delta v_x$.

Any curvilinear motion (which can't happen in 1-D motion) requires a sideways acceleration component of $$ a_{\perp} = \frac{v^2}{r} $$ where $v$ is the instantaneous speed and $r$ is the instantaneous radius of curvature.

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That is wrong. In uniform circular motion you have to use this equation: $$a=v^2/r $$ where $a$ is the acceleration and $v$ is the speed and $r$ is radius of the circular path. You have to know the radius of the circular path because an object with speed 5m/s will have different acceleration in circular paths with different radius. In short ,yes direction matter.

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You've confused velocity and speed. Speed is independent of direction, and is therefor always positive. For a 2-dimensional analysis, you have to consider the two different components of the velocity, Vx and Vy, but the speed S is $$S=\sqrt{V_x^2 + V_y^2}$$

Let's take a version of your example. An object travels along the x-axis at 5 m/s in the positive direction. 10 seconds later it is seen to be travelling parallel to the y-axis in the positive (y) direction. In the x-axis, the change in velocity was $$\Delta V_x = 0 - 5 = -5 m/s$$ and for the y-axis, $$\Delta V_y = 5 - 0 = 5 m/s$$ So the average acceleration over the 10 seconds for each axis is $$a_x =\frac {Delta V_x}{Delta t} = \frac {-5}{10} = -0.5 \frac{m}{s^2}$$ and $$a_y =\frac {Delta V_y}{Delta t} = \frac {5}{10} = 0.5 \frac{m}{s^2}$$

If the object had been given a negative y-velocity, the y-acceleration would have been negative.

While this is pretty obvious for right-angles, let's say the object changes course by 45 degrees, going up and to the right. Then the velocities are equal (since it's 45 degrees), and since the speed remains the same, $$5=\sqrt{V_x^2 + V_y^2} = \sqrt{2V_x^2}$$ and $$V_x = \frac{5}{sqrt{2}} = \frac{5}{1.414} = 3.54 \frac{m}{s}$$ and Vy is the same. You can now calculate the accelerations required.

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protected by Qmechanic Jan 11 '17 at 10:47

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