Question

A car drives at a constant speed around a banked circular track with a diameter of 136 m . The motion of the car can be described in a coordinate system with its origin at the center of the circle. At a particular instant the car's acceleration in the horizontal plane is given by a⃗ =(−15.4i^−25.4j^)m/s2.
What is the car's speed?
Where (x and y) is the car at this instant?

Answer 1

Answer:

$v = 44.9 m/s$

$x = 68 cos58.77 = 35.25 m$

$y = 68 sin58.77 = 58.14 m$

Explanation:

As we know that car drives at constant speed along circular path then we will have its position vector given as

$\vec r = R cos\omega t \hat i + R sin\omega t \hat j$

now if we differentiate is with respect to time then it will give as instantaneous velocity

so we have

$v = -R\omega sin\omega t \hat i + R\omega cos\omega t\hat j$

now again its differentiation with respect to time will give us acceleration

$a = - R \omega^2 cos\omega t \hat i - R\omega^2 sin\omega t\hat j$

now if we compare it with given value of acceleration

$a = -15.4\hat i - 25.4 \hat j$

$R\omega^2cos\omega t = 15.4$

$R\omega^2sin\omega t = 25.4$

divide both equations then we will have

[tec]tan\omega t = 1.65[/tex]

$\omega t = 58.77 degree$

Now we have

$R = 68 m/s$

so we can solve it for

$R\omega^2cos58.77 = 15.4$

$\omega = 0.66 rad/s$

so speed of the car is given as

$v = R\omega$

$v = 0.66\times 68$

$v = 44.9 m/s$

now we have coordinates of car given as

$x = R cos\omega t$

$x = 68 cos58.77 = 35.25 m$

$y = R sin\omega t$

$y = 68 sin58.77 = 58.14 m$

Answer 2

Answer:

speed = 44.9m/s

x = 35.5 m,  y = 58.0m

Explanation:

A car on a circular track with constant angular velocity ω can be described by the equation of position r:

$\overrightarrow {r(t)} = Rsin(\omega t)\hat{i} + Rcos(\omega t)\hat{j}$

The velocity v is given by:

$\overrightarrow {v(t)} = \overrightarrow{\frac{dr}{dt}}= \omega Rcos(\omega t)\hat{i} - \omega Rsin(\omega t)\hat{j}$

The acceleration a:

$\overrightarrow {a(t)} = \overrightarrow{\frac{dv}{dt}}= -\omega^2 Rsin(\omega t)\hat{i} - \omega^2 Rcos(\omega t)\hat{j}$

From the given values we get two equations:

$-\omega^2 Rsin(\omega t)=-15.4\\-\omega^2 Rcos(\omega t)=-25.4$

We also know:

$\overrightarrow {a(t)} = -\omega^2 Rsin(\omega t)\hat{i} - \omega^2 Rcos(\omega t)\hat{j}=-\omega^2\overrightarrow{r(t)}$

The magnitude of the acceleration a is:

$a=\sqrt{(-15.4)^2+(-25.4)^2}=29.7$

The magnitude of position r is:

$r=R=68m$

Plugging in to the equation for a(t):

$\overrightarrow {a(t)} =-\omega^2\overrightarrow{r(t)}$

and solving for ω:

$|\omega|=0.66$

Now solve for time t:

$\frac{sin(0.66t)}{cos(0.66t)}=tan(0.66t)=\frac{15.4}{25.4}\\t=0.83$

Using the calculated values to compute v(t):

$\overrightarrow {v(t)}= \omega Rcos(\omega t)\hat{i} - \omega Rsin(\omega t)\hat{j}\\\overrightarrow {v(t)}=44.88cos(0.55)\hai{i}-44.88sin(0.55)\hat{j}\\\overrightarrow {v(t)}=38.3\hat{i}-23.5\hat{j}$

The speed of the car is:

$\sqrt{38.3^2 + (-23.5)^2} = 44.9$

The position r:

$\overrightarrow {r(t)} = Rsin(\omega t)\hat{i} + Rcos(\omega t)\hat{j}\\\overrightarrow {r(t)} = 68sin(0.55)\hat{i} + 68cos(0.55)\hat{j}\\\overrightarrow {r(t)} = 35.5{i} + 58.0\hat{j}$