A car accelerates from rest at a rate of 6.50 m/s2. Determine thevelocity of the car at t = 4.00 s.
1 answer:
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Answer:
The rate of change of the shadow length of a person is 2.692 ft/s
Solution:
As per the question:
Height of a person, H = 20 ft
Height of a person, h = 7 ft
Rate = 5 ft/s
Now,
From Fig.1:
b = person's distance from the lamp post
a = shadow length
Also, from the similarity of the triangles, we can write:
Differentiating the above eqn w.r.t t:
Now, we know that:
Rate =
Thus
The velocity of B after elastic collision is 3.45m/s
This type of collision is an elastic collision and we can use a formula to solve this problem.
<h3>Elastic Collision</h3>The data given are;
- m1 = 281kg
- u1 = 2.82m/s
- m2 = 209kg
- u2 = -1.72m/s
- v1 = ?
Let's substitute the values into the equation.
From the calculation above, the final velocity of the car B after elastic collision is 3.45m/s.
Learn more about elastic collision here;