How Do Physics Formulas Apply to Airplane Acceleration and Projectile Motion?

In summary, the conversation revolved around solving various physics problems involving acceleration, velocity, and distance. The first question asked for the acceleration of an airplane given its initial speed and distance traveled. The second question involved finding the distance traveled by a fishing line thrown from a bridge with a given horizontal velocity. The last question asked for the resulting speed and direction of a boat moving relative to a current. The individual was seeking help understanding the necessary formulas and concepts to solve these problems.
  • #1
shortie
1
0
i have a few questions
1) an airplane is traveling at 140 km/h when it touches down. if it comes to a stop 1250m later what is the acceleration in m/s^2?
2)a bridge is 176.4 m above a river. if a fishing line is thrown from a bridge with a horizontal velocity of 22.0 m/s. how far will it have traveled when it hits the water?
and finally a boat is capable of moving 15.0m/s relative to the water. the captain sets a traveling dur north unaware that there is a current flowing from the east at 4.5 m/s. what is the resulatn speed and direction of the boat?

i don't need the answers i just need to know how to get them like the formulas n stuff
thx
 
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  • #2
Have you tried at all?

Do you have formulas like v(t)= v0+ at or
s(t)= (1/2)at2+ v0t+ x0?
 
  • #3


1) To calculate acceleration, we can use the formula a=(vf-vi)/t, where vf is the final velocity, vi is the initial velocity, and t is the time. In this case, the initial velocity is 140 km/h, which can be converted to 38.9 m/s. The final velocity is 0 m/s since the airplane comes to a stop. The distance traveled is 1250m. We can rearrange the formula to solve for acceleration, which gives us a=(0-38.9)/t. We need to find the time it takes for the airplane to come to a stop, which can be calculated using the formula d= vi*t + 1/2*a*t^2. Rearranging the formula gives us t=(vf-vi)/a. Plugging in the values, we get t=(0-38.9)/a. Now we can substitute this into the first formula to solve for acceleration, giving us a=(0-38.9)/((0-38.9)/a). This simplifies to a=-3.12 m/s^2.

2) To solve this problem, we can use the formula d=vi*t, where d is the distance, vi is the initial velocity, and t is the time. In this case, the initial velocity is 22.0 m/s and we need to find the time it takes for the fishing line to hit the water. The horizontal distance traveled is the same as the distance the fishing line travels, which is 176.4 m. Rearranging the formula, we get t=d/vi. Plugging in the values, we get t=176.4/22.0=8.02 seconds.

3) To find the resultant speed and direction of the boat, we can use vector addition. The boat is moving at a speed of 15.0 m/s north, and the current is flowing at a speed of 4.5 m/s east. To add these two vectors, we can use the Pythagorean theorem, which states that the magnitude of the resultant vector is equal to the square root of the sum of the squares of the individual vectors. In this case, the resultant speed is given by √(15.0^2 + 4.5^2) = 15.5 m/s. To find the direction of the resultant vector, we can use trigonometric functions
 

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