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

speed = 44.9m/s

x = 35.5 m, y = 58.0m

Step-by-step 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{(dr)/(dt)}= \omega Rcos(\omega t)\hat{i} - \omega Rsin(\omega t)\hat{j}`

The acceleration a:

`\overrightarrow {a(t)} = \overrightarrow{(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=√((-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:

`(sin(0.66t))/(cos(0.66t))=tan(0.66t)=(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:

`√(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}`

Answer 2

`v = 44.9 m/s`

`x = 68 cos58.77 = 35.25 m`

`y = 68 sin58.77 = 58.14 m`

Step-by-step 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* 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`