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?

Question

Physics

Salty

26 September, 22:52

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?

+3

Answers (1)

Know the Answer?

Amani · Answer 1

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°