Which of the following describes plastic? HELP
1 answer:
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First, create an illustration of the motion of the two cars as shown in the attached picture. The essential equations used are:
For constant acceleration:
a = v,final - v,initial /t
d = v,initial*t + 1/2*at²
For constant velocity:
d = constant velocity*time
The solutions is as follows:
a = v,final - v,initial /t
3.8 = (v₁ - 0)/4.6 s
v₁ = 17.48 m/s
Total distance = d1 + d2 + d3
d1 = d = v,initial*t + 1/2*at²
d2 = constant velocity*time
Total distance = 0*(4.6) + 1/2*(3.8)(4.6)² + (17.48)(9.2) + d3= 257.71
d3 = 56.69 m
For constant acceleration:
a = v,final - v,initial /t
d = v,initial*t + 1/2*at²
For constant velocity:
d = constant velocity*time
The solutions is as follows:
a = v,final - v,initial /t
3.8 = (v₁ - 0)/4.6 s
v₁ = 17.48 m/s
Total distance = d1 + d2 + d3
d1 = d = v,initial*t + 1/2*at²
d2 = constant velocity*time
Total distance = 0*(4.6) + 1/2*(3.8)(4.6)² + (17.48)(9.2) + d3= 257.71
d3 = 56.69 m
Answer:
t = 1.42 s and d = 35.5 m
Explanation:
Given that,
Velocity of a roadrunner is 25 m/s
A certain coyote wants to capture the roadrunner using a net dropped from an overpass that is 10 m high.
We need to find the time before the roadrunner is under the overpass and how far away from the overpass is the roadrunner when the coyote drops the net.
Let d is the distance traveled. So,
d = vt
d = 25 m/s × 1.42 s
d = 35.5 m