Q1. My brakes stop me at -5.5m/s^2 when I lock them up. A puppy runs in front of my car, and I stop just in time 3.5 seconds lat
1 answer:
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Answer:
a) Please, see the attched figure
b) Total displacement R = (78.3 km; -4.8 km)
c) R = (78.4 km * cos (-3.5°); 78.4 km * sin (-3.5°))
d) The hippo is 78.4 km from his starting point.
The total distance traveled is 102 km
Explanation:
a)Please, see the attached figure.
b) The vector A can be expressed as:
A = (magnitude * cos α; magnitude * sin α)
Where
magnitude = 42 km
α= 0
Then,
A = (42 km ; 0) or 42 km i
In the same way, we can proceed with the other vectors:
B = ( Bx ; By)
where
(apply trigonometry of right triangles: sen α = opposite / hypotenuse and
cos α = adjacent / hypotenuse. See the figure to determine which component of vector B is the opposite and adjacent side to α)
Bx = 28 km * sin 25 = 11.8 km
By = 28 km * cos 25 = -25.4 km (it has to be negative since it is directed towards the negative vertical region according to our reference system)
B = (11.8 km; -25.4 km) or 11.8 km i - 25.4 km j
C = (Cx; Cy)
where
Cx = 32 km * cos 40° = 24.5 km
Cy = 32 km * sin 40 = 20.6 km
C = (24.5 km; 20.6 km)
Then:
R = A+B+C = (42 km + 11.8 km + 24.5 km; 0 - 25.4 km + 20.6 km)
= (78.3 km; -4.8 km) or 78.3 km i -4.8 km j
c) R = (78.3 km; -4.8 km)
The magnitude of R is:
Using trigonometry, we can calculate the angle:
Knowing that
tan α = opposite / adjacent
and that
opposite = Ry = -4.8 km
adjacent = Rx = 78.3 km
Then:
tan α = -4.8 km / 78.4 km
α = -3.5°
We can now write the vector R in magnitude and direction form:
R = (78.4 km * cos (-3.5°); 78.4 km * sin (-3.5°))
d) The displacement of the hipo relative to the starting point is the magnitude of vector R calculated in c):
magnitude R = 78. 4 km
The total distance traveled is the sum of the magnitudes of each vector:
Total distance = 42 km + 28 km + 32 km = 102 km
Answer:
The distance the bungee cord that would be stretched 0.602 m, should be selected when pulled by a force of 380 N.
Explanation:
As from the given data
the length of the rope is given as l=30 m
the stretched length is given as l'=41m
the stretched length required is give as y=l'-l=41-30=11m
the mass is m=95 kg
the force is F=380 N
the gravitational acceleration is g=9.8 m/s2
The equation of k is given by equating the energy at the equilibrium point which is given as
Here
m=95 kg, g=9.8 m/s2, h=41 m, y=11 m so
Now the force is
or
So here F=380 N, k=630.92 N/m
So the distance is 0.602 m