Question

A car turns from a road into a parking lot and into an available parking space. The car’s initial velocity is 4 m/s [E 45° N]. The car’s velocity just before the driver decreases speed is 4 m/s [E 10° N]. The turn takes 3s. What's the average acceleration of the car during the turn? The answer should have directions with an angle.

Answer 1

Write the velocity vectors in component form.

• initial velocity:

v₁ = 4 m/s at 45º N of E

v₁ = (4 m/s) (cos(45º) i + sin(45º) j)

v₁ ≈ (2.83 m/s) i + (2.83 m/s) j

• final velocity:

v₂ = 4 m/s at 10º N of E

v₂ = (4 m/s) (cos(10º) i + sin(10º) j)

v₂ ≈ (3.94 m/s) i + (0.695 m/s) j

The average acceleration over this 3-second interval is then

a = (v₂ - v₁) / (3 s)

a ≈ (0.370 m/s²) + (-0.711 m/s²)

with magnitude

||a|| = √[(0.370 m/s²)² + (-0.711 m/s²)²] ≈ 0.802 m/s²

and direction θ such that

tan(θ) = (-0.711 m/s²) / (0.370 m/s²) ≈ -1.92

→   θ ≈ -62.5º

which corresponds to an angle of about 62.5º S of E, or 27.5º E of S. To use the notation in the question, you could say it's E 62.5º S or S 27.5º E.