Question

A car accelerates from rest at 5.75m/s2 for 4.4 sec when it runs out of pavement and runs into some mud. In the mud it accelerates to rest at -5.5m/s2. How far did it move from the beginning to the end? Give a variable legend for this problem. The model for this problem?

Answer 1

Answer:

113.85 m

Explanation:

When the car is on the pavement:

v₀ = 0 m/s

a = 5.75 m/s²

t = 4.4 s

Find: Δx and v

Δx = v₀ t + ½ at²

Δx = (0 m/s) (4.4 s) + ½ (5.75 m/s²) (4.4 s)²

Δx = 55.66 m

v = at + v₀

v = (5.75 m/s²) (4.4 s) + 0 m/s

v = 25.3 m/s

When the car is in the mud:

v₀ = 25.3 m/s

v = 0 m/s

a = -5.5 m/s²

Find: Δx

v² = v₀² + 2aΔx

(0 m/s)² = (25.3 m/s)² + 2 (-5.5 m/s²) Δx

Δx = 58.19 m

The total displacement is therefore:

55.66 m + 58.19 m = 113.85 m