Respuesta :
just use algebra to rearrange the terms so for s=1/2at^2 and solving for a we get 2s=at^2 so 2s/t^2=a
solving s=1/2at^2 for t gives 2s=at^2 which leads to 2s/a=t^2 so t=sqrt(2s/a)
filling in the numbers for 3 since a = 2s/t^2 we get a= (2(20.5m))/5s^2 =1.64m/s^2 the correct units in the answer indicate a possible correct answer since the units of acceleration are m/s^2.
4. using the value for a from question 3 and using 42.5 meters for s and solving for t we get sqrt((2s)/a) = sqrt((2*42.5m)/1.64m/s^2)= approximately 7.2s you can see the m's cancel and you are left s^2 as the units and the sqrt(s^2) = s so again the correct units indicate that 7.2s is probably correct. (These answers ARE correct though). checking the units gives a kinda quick assurance you are on the right path.
solving s=1/2at^2 for t gives 2s=at^2 which leads to 2s/a=t^2 so t=sqrt(2s/a)
filling in the numbers for 3 since a = 2s/t^2 we get a= (2(20.5m))/5s^2 =1.64m/s^2 the correct units in the answer indicate a possible correct answer since the units of acceleration are m/s^2.
4. using the value for a from question 3 and using 42.5 meters for s and solving for t we get sqrt((2s)/a) = sqrt((2*42.5m)/1.64m/s^2)= approximately 7.2s you can see the m's cancel and you are left s^2 as the units and the sqrt(s^2) = s so again the correct units indicate that 7.2s is probably correct. (These answers ARE correct though). checking the units gives a kinda quick assurance you are on the right path.
1) The formula for 'a' is a = 2S/t²
2) The formula for 't' is, t = √(2S/a)
3) The acceleration for the moon is a = 1.64 m/s²
4) It will take approximately 7 seconds for an object to fall 42.5 meters.
What is an equation?
"It is a statement which is consists of equal symbol between two mathematical expressions."
What is acceleration due to gravity?
"It is the acceleration gained by an object due to gravitational force."
For given example,
The formula S = 1/2at^2, represents the distance 's' that a free-falling object will fall near a planet or the moon in a given time 't' and 'a' represents the acceleration due to gravity.
1) to solve the formula for 'a'
⇒ S = 1/2 × a × t² ................(Given formula)
⇒ 2 × S = 2 × (1/2) × a × t² ..............(Multiply both the sides by 2)
⇒ 2S/t² = at²/t² ...............(Divide both the sides by t²)
⇒ 2S/t² = a
⇒ a = 2S/t²
Therefore, the formula for 'a' is a = 2S/t².
2) to solve the formula for 't':
⇒ S = 1/2 × a × t² ................(Given formula)
⇒ 2 × S = 2 × (1/2) × a × t² ..............(Multiply both the sides by 2)
⇒ 2S/a = at²/a ...............(Divide both the sides by a)
⇒ 2S/a = t²
⇒ t² = 2S/a
Take square-root on both the sides,
⇒ t = √(2S/a)
Since time t > 0, we take the positive square-root only.
Therefore, the formula for 't' is, t = √(2S/a)
3) Given: A free-falling object near the moon drops 20.5 meters in 5 seconds.
To find : the acceleration 'a' for the moon
Here, S = 20.5 meters and t = 5 seconds
Using the formula for 'a',
⇒ a = 2S/t²
⇒ a = (2 × 20.5)/(5)²
⇒ a = 1.64 m/s²
Therefore, the acceleration for the moon is a = 1.64 m/s².
4) Given: S = 42.5 meters
a = 1.64 m/s²
To find : time 't'
Using the formula for 't',
⇒ t = √(2S/a)
⇒ t = √ [(2 × 42.5) / 1.64]
⇒ t = 7.2 seconds
⇒ t ≈ 7 seconds
Therefore, it will take approximately 7 seconds for an object to fall 42.5 meters.
Learn more about equation here:
brainly.com/question/13170101
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