Question

1. A particle starts from rest, increasing its speed evenly, traveling a distance of 10.0 (m) in a time of 1.40 (s). From that moment it is kept at constant speed for 1.00 (s). Then its speed decreases to rest in a time of 2.80 (s). Calculate the total displacement of the particle.

STEP BY STEP PLEASE

Answer 1

Part 1

The particle starts from rest and travels a distance of 10 m in 1.4 sec.

Time period of the first part = T₁ = 1.4 sec

Distance traveled by the particle in the first part = D₁ = 10 m

Initial speed of the particle = V₁ = 0 m/s

Speed of the particle at the end of the first part = V₂

Acceleration of the particle in the first part = a₁

D₁ = V₁T₁ + a₁T₁²/2

10 = (0)(1.4) + a₁(1.4)²/2

a₁ = 10.2 m/s²

V₂ = V₁ + a₁T₁

V₂ = 0 + (10.2)(1.4)

V₂ = 14.28 m/s

Part 2

Particle travels at constant speed for 1 sec.

Time period of the second part = T₂ = 1 sec

Distance traveled by the particle in the second part = D₂

D₂ = V₂T₂

D₂ = (14.28)(1)

D₂ = 14.28 m

Part 3

Speed of the particle decreases to rest in a time of 2.8 sec.

Time period of the third part = T₃ = 2.8 sec

Final speed of the particle = V₃ = 0 m/s

Acceleration of the particle in the third part = a₃

V₃ = V₂ + a₃T₃

0 = 14.28 + a₃(2.8)

a₃ = -5.1 m/s²

Distance traveled by the particle in the third part = D₃

D₃ = V₂T₃ + a₃T₃²/2

D₃ = (14.28)(2.8) + (-5.1)(2.8)²/2

D₃ = 20 m

Total displacement of the particle = D

D = D₁ + D₂ + D₃

D = 10 + 14.28 + 20

D = 44.28 m

Total displacement of the particle = 44.28 m