Acar 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?

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:

The velocity v is given by:

The acceleration a:

From the given values we get two equations:

We also know:

The magnitude of the acceleration a is:

The magnitude of position r is:

Plugging in to the equation for a(t):

and solving for ω:

Now solve for time t:

Using the calculated values to compute v(t):

The speed of the car is:

The position r:

Explanation:

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

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

so we have

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

now if we compare it with given value of acceleration

divide both equations then we will have

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

Now we have

so we can solve it for

so speed of the car is given as

now we have coordinates of car given as

2b2t

explanation:

2b2t