Magnitude of velocity

Question

A car travels around a circular track with a radius of $r=250m$. When it is at point $A$ then $V_a=5m/s$ which increases at a rate of $\dot{v}=(0.06t)m/s$. Determine the magnitude of its velocity and its acceleration when it is $1/3$ around the track. My distance in this situation is obviously $524m$.

Asked this question earlier and while the answerer was extremely helpful the answer he gave me was incorrect or at least didn't give the answer my tutor is looking for. (I don't mean to appear ungrateful for the attempted help in my earlier question the gentleman was extremely helpful and nice.)

Help please

Answer 1

Hint: The acceleration would be the vector sum of the tangential and radial one, i.e $$|a|=\sqrt{(\frac{v^2}{r})^2+(\frac{dv}{dt})^2}$$ Now you have $a(t)=0.06t$, can you find $v(t)$? and subsequently $x(t)$? then you know $x=524m$, find t from this and then find v at that t and find |a| by the above relation.

Answer 2

Since the car undergoes pure circular motion ("pure" referring to the fact that it remains on a circular trajectory), we can specify its motion completely by the position along the path and the (tangential) speed.

The speed is given at a particular position (point A, which we will denote by $\theta = 0$) to be $v (t = 0) = v_0$, and is given to evolve according to $\dot{v} = f (t)$. Therefore, the speed at any given time is $v = v_0 + \int_0^t f (\tau) d \tau$. However, we would like to determine the car's speed (and the rate of change of speed) at a given position. To this end, we note that $s = R \theta$, where $s$ is the linear displacement along the track, $R$ is the radius of the track, and $\theta$ is the angular displacement. Therefore,

$$\theta = \frac{s}{R} = \frac{1}{R} \int_0^t v (\tau) d \tau$$

$$= \frac{v_0 t}{R} + \frac{1}{R} \int_0^t \int_0^\tau f (\tau') d \tau'$$

Letting $f (t) = K t$, we obtain that

$$\theta = \frac{v_0 t}{R} + \frac{K t^3}{6 R} = \Pi + \frac{1}{6} \frac{K R^2}{v_0^3} \Pi^3$$

where we define $\Pi = \frac{v_0 t}{R}$ as a non-dimensional time. This is a cubic equation that can be solved to obtain the time.

Now, we can find the speed of the car by noting that

$$v = v_0 + \frac{1}{2} K t^2 = v_0 \left( 1 + \frac{1}{2} \frac{K R^2}{v_0^3} \Pi^2 \right)$$

and the rate of change of speed of the car is

$$\dot{v} = K t = \frac{K R}{v_0} \Pi$$

The magnitude of the acceleration is found from the vectorial sum of the radial and tangential components of acceleration:

$$|\vec{a}| = \left| \left\langle \frac{v^2}{R}, \dot{v} \right\rangle \right|$$

$$= \frac{v_0^2}{R} \left| \left\langle 1 + \frac{1}{2} \frac{K R^2}{v_0^3} \Pi^2, \frac{K R^2}{v_0^3} \Pi \right\rangle \right|$$

$$= \frac{v_0^2}{R} \sqrt{1 + \left( \frac{K R^2}{v_0^3} + \frac{K^2 R^4}{v_0^6} \right) \Pi^2 + \frac{1}{4} \frac{K^2 R^4}{v_0^6} \Pi^4}$$

For the given example, with $v_0 = 5 \frac{\text{m}}{\text{s}}$, $R = 250 \text{ m}$, and $K = 0.06 \frac{\text{m}}{\text{s}^3}$, the non-dimensional parameter $\frac{K R^2}{v_0^3} = 30$, and the only real solution to

$$\theta = \frac{2}{3} \pi = \Pi + \frac{1}{6} (30) \Pi^3$$

is $\Pi \approx 0.6596$.

Although not necessary to calculate for this problem, this corresponds to a time of

$$t = \frac{R}{v_0} \Pi$$

$$\approx (50 \text{ s}) (0.6596) = 33.0 \text{ s}$$

The speed (magnitude of velocity) at this instant is

$$v = v_0 \left( 1 + \frac{1}{2} (30) (0.6596)^2 \right)$$

$$\approx \left( 5 \frac{\text{m}}{\text{s}} \right) (7.53) = 37.6 \frac{\text{m}}{\text{s}}$$

The magnitude of acceleration at this instant is

$$a = \frac{v_0^2}{R} \sqrt{1 + (30 + 30^2) (0.6596)^2 + \frac{1}{4} (30)^2 (0.6596)^4}$$

$$\approx \left( 0.1 \frac{\text{m}}{\text{s}^2} \right) (21.2) = 2.12 \frac{\text{m}}{\text{s}^2}$$