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 July 14th, 2016, 03:25 PM #1 Newbie   Joined: Jul 2016 From: Brazil Posts: 1 Thanks: 0 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;
 July 14th, 2016, 04:46 PM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,772 Thanks: 1426 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) ... $a_c = \dfrac{v_0^2}{r} = r \cdot \omega_0^2$ Thanks from topsquark and manus
July 14th, 2016, 07:16 PM   #3
Math Team

Joined: Jul 2011
From: Texas

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Thanks: 1426

Quote:
 Originally Posted by skeeter 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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