Question

A car starts from rest and moves around a circular track of radius 37.0 m. Its speed increases at the constant rate of 0.440 m/s2. (a) What is the magnitude of its net linear acceleration 18.0 s later? (b) What angle does this net acceleration vector make with the car's velocity at this time?

Answer 1

Answer:

a) 1.76 m/s²

b) 75.5°

Explanation:

1. linear(tangential) acceleration a:

a = 0.44m/s²

2. speed v:

v(t) = at

3. radial acceleration α on a track with radius r:

α(t) = v²/r = (at)²/r

α(18)= (0.44*18)²/37 = 1.7 m/s²

4. total acceleration |a|:

|a| = √(a²+α²)=1.76m/s²

5. angle Ф with v:

tanФ =(α/a) => Ф = 75.5°