Question

A mobile starts and travels 225 m with an acceleration of (- 3.5i + 4.5j) 2 along a rectilinear path. Determine:

a) The time spent
b) The final speed
c) The average speed I hope you can help me.

Answer 1

a) 8.89 s

b) 50.6 m/s

c) 25.3 m/s

Step-by-step explanation:

d = 225 m

v₀ = 0 m/s

a = (-3.5i + 4.5j) m/s²

First, find the magnitude of the acceleration.

a = √((-3.5)² + (4.5)²) m/s²

a = 5.7 m/s²

a) Find t

Δx = v₀ t + ½ at²

225 m = (0 m/s) t + ½ (5.7 m/s²) t²

t = 8.89 s

b) Find v

v² = v₀² + 2aΔx

v² = (0 m/s)² + 2 (5.7 m/s²) (225 m)

v = 50.6 m/s

c) Find v_avg

v_avg = Δx / t

v_avg = 225 m / 8.89 s

v_avg = 25.3 m/s

Or, for constant acceleration:

v_avg = (v₀ + v) / 2

v_avg = (0 m/s + 50.6 m/s) / 2

v_avg = 25.3 m/s