Answer: the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m
Explanation:
Given that;
mass of vehicle m = 1000 kg
for a low speed test; V = 2.5 m/s
bumper maximum deflection = 4 cm = 0.04 m
First we determine the energy of the vehicle just prior to impact;
W_v = 1/2mv²
we substitute
W_v = 1/2 × 1000 × (2.5)²
W_v = 3125 J
now, the the effective design stiffness k will be:
at the impact point, energy of the vehicle converts to elastic potential energy of the bumper;
hence;
W_v = 1/2kx²
we substitute
3125 = 1/2 × k (0.04)²
3125 = 0.0008k
k = 3125 / 0.0008
k = 3906250 N/m
Therefore, the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m
6x2=12m
6x18=108
12m+108
Simplified: m+9 bc 12/12 and 108/12
Answer:1) the total distance is the sum of the two distances
60 km + 45 km = 105 km
2) The displacement is the net movement, or the difference between the initial position and the final position
Call x the initial position, then the final position is x + [60km - 45km]
And the displacement is x + (60km - 45km) - x =60km -45 km = 15 km
Explanation:
