a race car starts from rest on a circular track.the car increases its speed at constant rate at as it goes 2.75 times around the track.find the angle that the total acceleration of the car makes with the radius connecting the centre of the track and the car at the moment the car completes its trip of 2.75 times around the circle.

Answer 1

Let a_t be the tangential acceleration and a_c be the centripetal acceleration.
Let v be the final velocity, theta be the angle that a_t makes with the radius, and fi be the angle that a_c makes with the radius.
Let r be the radius of the track, and x be the distance traveled by the car.

The car goes 2.75 times around the track, so x = 2.75 * 2 * pi * r.

We can write down two equations for the tangential acceleration a_t along the track:
a_t = v^2 / (2 * x)
a_t * path time = v

This yields three equations for three unknowns a_t, v, and x:
a_t = v^2 / (2*2.75*2*pi*r)
a_t * 2.75 * 2 * pi = v
x = 2.75*2*pi*r

Let's express v in terms of a_t from the second equation:
v = a_t * 2.75 * 2 * pi

Now let's substitute this in the first equation:
a_t = (a_t * 2.75 * 2 * pi)^2 / (2*2.75*2*pi*r)

We can see that a_t terms are cancelled out, and we can solve for r:
r = (2.75 * 2 * pi)^2 / (2*2.75*2*pi) = 2.75 * pi

Now we can find the centripetal acceleration a_c using the final velocity v and r:
a_c = v^2 / r = (a_t * 2.75 * 2 * pi)^2 / (2.75 * pi)

To find the angle between the total acceleration and the radius, we can use the tangent of the angle θ:
tan(θ) = a_c / a_t = ((a_t * 2.75 * 2 * pi)^2 / (2.75 * pi)) / a_t

We can cancel out the a_t terms:
tan(θ) = (2.75 * 2 * pi)^2 / (2.75 * pi)

Now we can find the angle θ:
θ = atan(tan(θ)) = atan((2.75 * 2 * pi)^2 / (2.75 * pi))

θ ≈ 20.6 degrees

So the angle that the total acceleration makes with the radius connecting the center of the track and the car is approximately 20.6 degrees.