- #1
Bunny-chan
- 105
- 4
There is a problem in my Physics textbook which says:
"A car runs counter-clockwise in a circular lane of [itex]1 km[/itex] of diameter, going through the south extreme at [itex]60 km/h[/itex] on the instant [itex]t = 0[/itex]. From that point onwards, the driver accelerates the car uniformely, reaching [itex]240 km/h[/itex] in [itex]10[/itex] seconds.
Determine the vector average acceleration between [itex]t = 0[/itex] and [itex]t = 10s[/itex]"
P.S: Sorry for any typos, I had to translate it from a textbook written in Portuguese and I'm not too familiar with english technical language yet. Anyways...
3. The Attempt at a Solution [/B]
So what we know:
Diameter of the circunference: [itex]1km[/itex];
Radius of the circunference: [itex]\frac{D}{2} = 500m[/itex];
Initial position: [itex]0[/itex];
Initial velocity: [itex]60km/h[/itex] or [itex]16.7m/s[/itex];
Velocity after [itex]10s[/itex]: [itex]240km/h[/itex] or [itex]66.7m/s[/itex];
Acceleration in the interval: [tex]\vec a = \frac{\Delta \vec{V}}{\Delta t} = \frac{66.7-16.7}{10-0} = \frac{50}{10} = 5m/s^2 [/tex]Total displacement: [tex]S = S_0 + \vec{V_0}t + \frac{at^2}{2} \Rightarrow S = 0 + 16.7 \times 10 + \frac{5 \times 10^2}{2} = 167 + 250 = 417m[/tex]Now, vectors contain sense, magnitude and direction, so, as it is a circular motion problem, we should find out the angle of the arc inscribed by the displacement:[tex]\theta = \frac{360^\circ}{2\pi r} \times 417 \Rightarrow \theta = \frac{360^\circ}{2\pi \times 500} \times 417 = \frac{360^\circ}{1000 \pi} \times 417 \simeq 47.8^\circ[/tex]Now, I'm not really sure what to do next, but I've tried Law of Cosines:[tex]\vec{V_r}^2 = \vec{V_1}^2 + \vec {V_2}^2 - 2 \vec{V_1} \vec{V_2} \times \cos A \Rightarrow \vec{V_r}^2 = 16.7^2 + 66.7^2 - 2 \times 16.7 \times 66.7 \times \cos 47.8 = 278.89 + 4448.89 - 1496.4 = 3231.38 \Rightarrow \vec{V_r} = \sqrt{3231.38} \simeq 56.8m/s[/tex]Continuing:[tex]\vec{a_avg} = \frac{\Delta \vec{V}}{\Delta t} = \frac{56.8}{10} = 5.68m/s^2[/tex]OK. So that matches the answer in my textbook, in regards to the magnitude. But the problem is that my answer says nothing about the direction and the sense of the vector, which should be [itex]60.3 ^\circ[/itex] north from direction east, according to my textbook.
And now I have no idea about what to do to be able to calculate that. My understanding of the circular motion is still quite blurry, and I have some other questions too:
1) How was I able to solve this problem using a linear movement formula for displacement and acceleration? Why didn't I need to use the angular formulas and all of that?
2) I couldn't grasp what velocity and acceleration meant in this problem. Were they both tangencial? Was acceleration centripetal...?
I've invested some time writing this so if someone could help me I'd be really grateful. I have nobody else to ask. x_x
Homework Statement
"A car runs counter-clockwise in a circular lane of [itex]1 km[/itex] of diameter, going through the south extreme at [itex]60 km/h[/itex] on the instant [itex]t = 0[/itex]. From that point onwards, the driver accelerates the car uniformely, reaching [itex]240 km/h[/itex] in [itex]10[/itex] seconds.
Determine the vector average acceleration between [itex]t = 0[/itex] and [itex]t = 10s[/itex]"
P.S: Sorry for any typos, I had to translate it from a textbook written in Portuguese and I'm not too familiar with english technical language yet. Anyways...
Homework Equations
3. The Attempt at a Solution [/B]
So what we know:
Diameter of the circunference: [itex]1km[/itex];
Radius of the circunference: [itex]\frac{D}{2} = 500m[/itex];
Initial position: [itex]0[/itex];
Initial velocity: [itex]60km/h[/itex] or [itex]16.7m/s[/itex];
Velocity after [itex]10s[/itex]: [itex]240km/h[/itex] or [itex]66.7m/s[/itex];
Acceleration in the interval: [tex]\vec a = \frac{\Delta \vec{V}}{\Delta t} = \frac{66.7-16.7}{10-0} = \frac{50}{10} = 5m/s^2 [/tex]Total displacement: [tex]S = S_0 + \vec{V_0}t + \frac{at^2}{2} \Rightarrow S = 0 + 16.7 \times 10 + \frac{5 \times 10^2}{2} = 167 + 250 = 417m[/tex]Now, vectors contain sense, magnitude and direction, so, as it is a circular motion problem, we should find out the angle of the arc inscribed by the displacement:[tex]\theta = \frac{360^\circ}{2\pi r} \times 417 \Rightarrow \theta = \frac{360^\circ}{2\pi \times 500} \times 417 = \frac{360^\circ}{1000 \pi} \times 417 \simeq 47.8^\circ[/tex]Now, I'm not really sure what to do next, but I've tried Law of Cosines:[tex]\vec{V_r}^2 = \vec{V_1}^2 + \vec {V_2}^2 - 2 \vec{V_1} \vec{V_2} \times \cos A \Rightarrow \vec{V_r}^2 = 16.7^2 + 66.7^2 - 2 \times 16.7 \times 66.7 \times \cos 47.8 = 278.89 + 4448.89 - 1496.4 = 3231.38 \Rightarrow \vec{V_r} = \sqrt{3231.38} \simeq 56.8m/s[/tex]Continuing:[tex]\vec{a_avg} = \frac{\Delta \vec{V}}{\Delta t} = \frac{56.8}{10} = 5.68m/s^2[/tex]OK. So that matches the answer in my textbook, in regards to the magnitude. But the problem is that my answer says nothing about the direction and the sense of the vector, which should be [itex]60.3 ^\circ[/itex] north from direction east, according to my textbook.
And now I have no idea about what to do to be able to calculate that. My understanding of the circular motion is still quite blurry, and I have some other questions too:
1) How was I able to solve this problem using a linear movement formula for displacement and acceleration? Why didn't I need to use the angular formulas and all of that?
2) I couldn't grasp what velocity and acceleration meant in this problem. Were they both tangencial? Was acceleration centripetal...?
I've invested some time writing this so if someone could help me I'd be really grateful. I have nobody else to ask. x_x
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