Answer:
a = (v2 - v1) / t
From A to B (8 - 4) m/s / 1 s = 4 m / s^2
From A to D ( 7 - 4) m/s / 5 s = .6 m / s^2
Note these equations hold for "uniform" values
They say nothing about the acceleration at intermediate points - the equation just says that his average speed increased from 4 m/s to 7 m/s during a 5 sec period
The final velocity of the red barge in the collision elastic is 0.311 m/s when it collides with blue barge pf mass 1000000 kg.
Final velocity(v3) of the red barge is calculated by following formula
m1×v1+ m2×v2= (m1+m2)v3
Substituting the value of m1= 150000 kg, v1= 0.25 m/s, m2= 1000000 kg, v2= 0.32 m/s
150000 × 0.25+ 1000000×0.32= (150000+1000000)×v3
37500+ 320000= 1150000×v3
357500= 1150000×v3
v3= 0.311 m/s
<h3>What is elastic collision velocity? </h3>
- The velocity of the target particle after a head-on elastic impact in which the projectile is significantly more massive than the target will be roughly double that of the projectile, but the projectile velocity will remain virtually unaltered.
For more information on elastic collision velocity kindly visit to
brainly.com/question/29051562
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Here, we know, according to 3rd Equation of Kinematics,
v² - u² = 2as
Here, u = 0 [ Free fall ]
a = 9.8 m/s² [ constant value for the Earth system ]
s = 15 m
Substitute their values,
v² - 0² = 2 * 9.8 * 15
v² = 294
v = √294
v = 17.15 m/s
In short, Your Answer would be Option D
Hope this helps!
