Answer:
v = 21.4m/s
Explanation:
Given r1 = < 13, -18, -2 > m,
F = < 250, 390, -220 > N, r2 = < 19, -23, -5 > m
Where r1 and r2 are position vectors in space and F is the constant force vector in space.
r2 - r1 = < 19–13, -23–(-18), -5–(-2) > m
Δr = < 6, -5, -3 > m
F = m×a
a = F/m
m = 110kg (scalar quantity)
a = < 250/110, 390/110, -220/110 >
a = < 2.27, 3.55, -2 > m/s²
Since the force is constant, the acceleration is also constant. Therefore the equations of constant acceleration motion apply here.
V² = u² +2aΔr
The magnitude of the acceleration is calculated as follows
a = √(2.27² + 3.55² +(-2)²)
a = √(21.7554)
a = 4.66m/s²
Magnitude of Δr,
Δr = √(6² + (-5)² +(-3)²) = √(70) = 8.37m
Having calculated the magnitude of the acceleration and displacement, we can now calculate the final velocity.
u = initial velocity = 19.5m/s
v² = 19.5² + 2×4.66×8.37
v² = 458.25
v = √(458.25)
v = 21.4m/s
