Effect of Mass and Force on Acceleration and Distance Traveled in Kinematics

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SUMMARY

The discussion focuses on the effects of mass and force on acceleration and distance traveled in kinematics. When the mass of a body is doubled and the force applied is increased five-fold, the acceleration increases by a factor of 2.5, as derived from the equation xa = 5f/2m. Consequently, the distance traveled in a given time also increases by 2.5 times, confirmed through the kinematic equation d = 0.5at². The consistent factor of 0.5 in the distance equation remains unchanged, reinforcing the calculations presented.

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[Solved] Kinematics ratio problem

Homework Statement


A force F produces an acceleration a when applied to a certain body. If the mass of the body is doubled and the force increased five-fold, what will be the effect on the following?
a. its acceleration

b. the distance the body travels in a given time


Homework Equations





The Attempt at a Solution


a. i know its acceleration is increased by 2.5 because if we have xa = 5f/2m, x=5/2

b. i know the distance is increased by 2.5, because everything else remains unchanged... but i can't seem to explain it in using equations... d = v1*t + 1/2at^2, if everything remains unchanged except a, we get: xd = 1/2(5/2a)1t^2 (assuming your velocity is 0 m/s?) i get that x = 5/4 but i know that's wrong, the distance can't be increased by 5/4... :confused:
 
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Karajovic said:

Homework Statement


A force F produces an acceleration a when applied to a certain body. If the mass of the body is doubled and the force increased five-fold, what will be the effect on the following?
a. its acceleration

b. the distance the body travels in a given time


Homework Equations





The Attempt at a Solution


a. i know its acceleration is increased by 2.5 because if we have xa = 5f/2m, x=5/2

b. i know the distance is increased by 2.5, because everything else remains unchanged... but i can't seem to explain it in using equations... d = v1*t + 1/2at^2, if everything remains unchanged except a, we get: xd = 1/2(5/2a)1t^2 (assuming your velocity is 0 m/s?) i get that x = 5/4 but i know that's wrong, the distance can't be increased by 5/4... :confused:

Remember that the 1/2 factor is there regardless if the acceleration is changed or not.

d1 = 0.5at2
d2 = 0.5(2.5a)t2 = 2.5(0.5at2) = 2.5d1
 
danago said:
Remember that the 1/2 factor is there regardless if the acceleration is changed or not.

d1 = 0.5at2
d2 = 0.5(2.5a)t2 = 2.5(0.5at2) = 2.5d1

Thank you very much!
 

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