A race car starts from rest on a circular track. The car increases its speed at a constant rate at as it goes 4.25 times around the track. Find the angle that the total acceleration of the car makes with the radius connecting the center of the track and the car at the moment the car completes its trip of 4.25 times around the circle.

Now, Tangential and radial acceleration are always perpendicular to each other. Thus, in a triangle system they are opposite and adjacent sides. From trigonometric ratios,

Opposite/Adjacent = tanθ

Now centripetal acceleration is given as v²/r

Thus, the angle to the radial is given as;

tanθ = translational acceleration/centripetal acceleration

So, tanθ = a/(v²/r) = ar/v²

Thus, θ = tan^(-1)(ar/v²)

Now, Distance of one round of circular motion of 'r' radius

= circumference of circle = 2πr

N = number of trips car makes around the circle

Thus,

total distance = 2πrN

Now, from equation of motion, we know that v² = u² + 2as

Where s is total distance and u is initial velocity which is zero in this case.

Thus, v² = 0² + 2as

Making s the subject;

s = v²/2a

Thus,

2πrN = v²/2a

So let's simplify to bring out ar/v² which is what we will use to calculate the angle. So,

2πrN x 2a = v²

4πN(ar) = v²

Thus, ar/v² = 1/(4πN)

From the question, N = 4.25

Thus,

ar/v² = 1/(4π x 4.25) = 0.0187

From earlier, we saw that the angle is given by;

θ = tan^(-1)(ar/v²)

Thus, θ = tan^(-1)(0.0187)

θ = 1.07°