Question

The car travels along a road which for a short distance is defined by r= (200/θ) m, where θ is in radians. If it maintains a constant speed of v= 35 m/s, determine the radial and transverse components of its velocity when θ= /3 dians

Answer 1

Answer:

Transverse speed of the car is

$v_t = 25.2 m/s$

Now radial velocity given as

$v_r = 24.1 m/s$

Explanation:

As we know that the tangential displacement of the object is given as

$ds = R d\theta$

now its velocity in tangential direction is given as

$v = \frac{ds}{dt}$

so we have

$v_t = R\frac{d\theta}{dt}$

$v_t = (\frac{200}{\theta}).\frac{d\theta}{dt}$

also in radial direction the speed is given as

$v_r = \frac{dR}{dt}$

$v_r = -\frac{200}{\theta^2}.\frac{d\theta}{dt}$

we know that total speed of the object is constant so we have

$v_r^2 + v_t^2 = v^2$

$(\frac{200}{\theta}\frac{d\theta}{dt})^2 + (-\frac{200}{\theta^2}.\frac{d\theta}{dt})^2 = 35^2$

$\frac{d\theta}{dt} = \frac{35}{\sqrt{(\frac{200}{\theta})^2 + (\frac{200}{\theta^2})^2}$

so we have

$\theta = \frac{\pi}{3}$

$\frac{d\theta}{dt} = 0.132 rad/s$

now we have transverse velocity given as

$v_t = (\frac{200}{\pi/3}).0.132$

$v_t = 25.2 m/s$

Now radial velocity given as

$v_r = -\frac{200}{(\pi/3)^2}.0.132$

$v_r = -24.1 m/s$

#Learn

Topic : Kinematics

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