A horizontal force of 150 N is used to push a 38.0 kg packing crate a distance of 6.85 m on a rough horizontal surface. If the c
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Answer:
his acceleration rate is -0.00186 m/s²
Explanation:
Given;
initial position of the car, x₀ = 100 miles = 160, 900 m ( 1 mile = 1609 m)
time of motion, t₀ = 60 minutes = 60 mins x 60 s = 3,600 s
final position of the car, x₁ = 150 miles = 241,350 m
time of motion, t₁ = 100 minutes = 100 mins x 60 s = 6,000 s
The initial velocity is calculated as;
u = 160, 900 m / 3,600 s
u = 44.694 m/s
The final velocity is calculated as;
v = 241,350 m / 6,000 s
v = 40.225 m/s
The acceleration is calculated as;
Therefore, his acceleration rate is -0.00186 m/s²
Answer:
The dog catches up with the man 6.1714m later.
Explanation:
The first thing to take into account is the speed formula. It is , where v is speed, d is distance and t is time. From this formula, we can get the distance formula by finding d, it is
Now, the distance equation for the man would be:
The distance equation for the dog would be obtained by the same way with just a little detail. The dog takes off running 1.8s after the man did. So, in the equation we must subtract 1.8 from t.
For a better understanding, at t=1.8 the dog must be in d=0. Let's verify:
Now, for finding how far they have each traveled when the dog catches up with the man we must match the equations of each one.
The result obtained previously means that the dog catches up with the man 3.8571s after the man started running.
That value is used in the man's distance equation.
Finally, the dog catches up with the man 6.1714m later.