On a speed-versus-time graph, a straight line that slopes down toward
the right shows that the object is experiencing a constant deceleration.
When they meet the 40-kg boy will have moved a distance of 6 m.
Displacement of the 40 kg boy
The displacement of the 40 kg boy is calculated from the principle of center mass.
X(40 kg) = (60 x 10 m + 40 x 0)/(40 kg + 60 kg)
X(40 kg) = (600)/(100) = 6 m
X(60 kg) = (60 x 0 + 40 x 10 m)/(40 kg + 60 kg)
X(60 kg) = (400)/(100) = 4 m
Thus, when they meet the 40-kg boy will have moved a distance of 6 m.
Learn more about center mass here: brainly.com/question/13499822
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One thing you should notice. It is kind of weird. Ke has no direction so that fact that it has velocities associated with it does not matter. It becomes a scaler (something measured by amount alone).
General Formula
Ke = 1/2 m v^2
Formula for this problem
Ke = 1/2 m (v2)^2 - 1/2 m (v1)^2
Givens
m = 1200 kg
v2 = 100 km/hr = 100 km/h * [1 hour / 3600 sec] * [1000 m/ 1km] = 27.8 m/s
v1 = 50 km / hr = 13.9 m/s
Substitution and work.
================
delta Ke = 1/2 1200 (27.8)^2 - 1/2 1200 (13.9)^2
delta Ke = 463704 - 115926
delta Ke = 34778 Joules
delta Ke = 34.8 kJ
The change is 34.8 kJ which means that the vehicle gains 34.8 kJ
Answer:
0.055 * 5.1 = 0.2805
Explanation:
I used a calculator so it should be accurate! yw!! :D
