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July 14th, 2016, 03:25 PM   #1
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Question physics calculation, help me please!

A car starts a curve with speed 50km / h. This curve is an arc of circumference with radius of 10 m. If it takes 10 seconds to describe

this curve, and is a constant acceleration of 0.8 m / s, calculate:
a) the angular velocity curve beginning and at the end thereof;
b) Distance traveled this curve;
c) Angle described the car;
d) angular acceleration;
e) centripetal acceleration at the beginning of the path;
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July 14th, 2016, 04:46 PM   #2
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first, you need to convert the initial tangential velocity, $v_{0} = 50 \, km/hr$ to $m/s$

(a) ...

$\omega_0 = r \cdot v_{0}$

assuming the given constant acceleration is tangential to the curve ...

$v_{f} = v_{0} + a_T \cdot \Delta t$ and $\omega_f = r \cdot v_{f}$

(b) ...

$\Delta s = r \cdot \bar{\omega} \cdot \Delta t = r \left(\dfrac{\omega_0+\omega_f}{2}\right) \cdot \Delta t$

(c) ...

$\Delta \theta = \dfrac{\Delta s}{r}$

(d) ...

$\alpha = \dfrac{\Delta \omega}{\Delta t}$

(e) ...

$a_c = \dfrac{v_0^2}{r} = r \cdot \omega_0^2$
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July 14th, 2016, 07:16 PM   #3
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Quote:
Originally Posted by skeeter View Post
first, you need to convert the initial tangential velocity, $v_{0} = 50 \, km/hr$ to $m/s$

(a) ...

$\omega_0 = \dfrac{v_{0}}{r}$

assuming the given constant acceleration is tangential to the curve ...

$v_{f} = v_{0} + a_T \cdot \Delta t$ and $\omega_f = \dfrac{v_{f}}{r}$

(b) ...

$\Delta s = r \cdot \bar{\omega} \cdot \Delta t = r \left(\dfrac{\omega_0+\omega_f}{2}\right) \cdot \Delta t$

(c) ...

$\Delta \theta = \dfrac{\Delta s}{r}$

(d) ...

$\alpha = \dfrac{\Delta \omega}{\Delta t}$

(e) ...

$a_c = \dfrac{v_0^2}{r} = r \cdot \omega_0^2$
couple of corrections for part (a) ... sorry
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