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
Answer:
<h2>7.5 N</h2>
Explanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question
mass = 2.5 kg
acceleration = 3.0 m/s²
We have
force = 2.5 × 3.0 = 7.5
We have the final answer as
<h3>7.5 N</h3>
Hope this helps you
Answer:
3secs
Explanation:
Given the following parameters
height H= 81.3m
Velocity v = 12.4m/s
Required
Time it take to reach the ground
Using the equation of motion
H = ut+1/2gt²
81.3 = 12.4t + 1/2(9.8)t²
81.3 = 12.4t + 4.9t²
4.9t² + 12.4t - 81.3 = 0
Using the general formula to find t
t = -12.4±√12.4²-4(4.9)(-81.3)/2(4.9)
t = -12.4±√153.76+1593.48/2(4.9)
t = -12.4±√1747.24/9.8
t = -12.4+41.8/9.8
t = 29.4/9.8
t = 3secs
Hence it took 3secs to reach the ground
Answer:
60,000m
Explanation:
Convert km/h to m/s by multiplying with 1000/3600.
Convert hours to seconds by multiplying with 3600.
Because displacement is a vector quantity and deals with the shortest distance between points, simply plug it into the equation s=vt.
Answer:
2m/s²
Explanation:
velocity = displacement (distance in a specified direction /time
