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Sedbober [7]
3 years ago
10

A hiker travels south along a straight line path for 1.5 hours with an average velocity of 0.75 km/hr, then travels south for 2.

5 hours with an average velocity of 0.90 km/hr. what is the hiker's displacement for the total trip?
Physics
1 answer:
ANTONII [103]3 years ago
7 0
Distance travelled in south direction= 1.5hr*0.75km/hr= 1.125km
Distance travlled in north direction= 0.90*2.5=2.25
Net displacement = 2.25-1.125= 1.125 to the north
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grin007 [14]

Answer: 17.68 s

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:  

y=y_{o}+V_{o}t-\frac{1}{2}gt^{2} (1)  

Where:  

y=0 is the height of the ball when it hits the ground  

y_{o}=70 m is the initial height of the ball

V_{o}=82m/s is the initial velocity of the ball  

t is the time when the ball strikes the ground

g=9.8m/s^{2} is the acceleration due to gravity  

Having this clear, let's find t from (1):  

0=70m+(82m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2} (2)  

Rewritting (2):

-\frac{1}{2}(9.8m/s^{2})t^{2}+(82m/s)t+70m=0 (3)  

This is a quadratic equation (also called equation of the second degree) of the form at^{2}+bt+c=0, which can be solved with the following formula:

t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}  (4)

Where:

a=-\frac{1}{2}(9.8m/s^{2}

b=82m/s

c=70m

Substituting the known values:

t=\frac{-82 \pm \sqrt{82^{2}-4(-\frac{1}{2}(9.8)(70)}}{2a}  (5)

Solving (5) we find the positive result is:

t=17.68 s

7 0
3 years ago
Find the hiker’s gravitational potential energy if the cliff is 60m high
Furkat [3]

Answer:

Potential energy is U=mgh

Explanation:

The potential energy depends on the mass, the acceleration of gravity g and the height at which the object or person is.

Potential energy  U=mgh

In this case we would need to know the exact mass of the hiker in order to calculate the potential energy.

But we know the values of g and h

g=9.81m/s^2

h=60m

So, the potential energy

U=m(9.81m/s^2)(60m)\\\\U=588.6*m

m is the mass of the hiker, wich is not in the description of the problem.

4 0
3 years ago
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hoa [83]

Refer to the attachment

5 0
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The length of a rectangular sheet of metal decreases by 34.5 cm. Its width decreases proportionally. If the sheets original widt
devlian [24]
The original width was 94.71 cm 
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