MCAT physics q - free fall.

[AG22]

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Hey everyone,

A question describes how two balls of different materials are dropped from a height and air resistance is neglible. If an object falls a distance of delta x during the first t seconds, how far does it fall during the first 3t seconds?

I realised that the object is uniformly accelarating (i.e. gravity) Soo, i used this equation: delta x = (v avg)(delta t) [because when there is a uniform acceleration, v avg = (v1+v2)/2]
That means that the object will fall 3 times delta x. Right? since x is proportional to t?

The solutions use the formula delta x = v1t +1/2at^2 And since v1 = 0, delta x is proportional to t^2. SO the real answer is the object will fall 9 times delta x in the first 3t seconds.

Can any one explain to me why I can't use this equation delta x = (v avg)(delta t) and hence cannot use the fact that x is proportional to t to give me the right answer? (even when it is uniformly accelerating?)

It would be greatly appreciated!

 

[Bernoull]

Hey everyone,

A question describes how two balls of different materials are dropped from a height and air resistance is neglible. If an object falls a distance of delta x during the first t seconds, how far does it fall during the first 3t seconds?

I realised that the object is uniformly accelarating (i.e. gravity) Soo, i used this equation: delta x = (v avg)(delta t) [because when there is a uniform acceleration, v avg = (v1+v2)/2]
That means that the object will fall 3 times delta x. Right? since x is proportional to t?

The solutions use the formula delta x = v1t +1/2at^2 And since v1 = 0, delta x is proportional to t^2. SO the real answer is the object will fall 9 times delta x in the first 3t seconds.

Can any one explain to me why I can't use this equation delta x = (v avg)(delta t) and hence cannot use the fact that x is proportional to t to give me the right answer? (even when it is uniformly accelerating?)

It would be greatly appreciated!

Your intuition and reasoning are correct. If acceleration is constant, dx= Vavg*dt. Both formulae are correct, you can derive
dx=xo + vot +0.5 (at^2) with calculus. I think your mistake stems from ur calculation of Vavg:

I think the problem asks by what factor does the object travel in 3t secs compared to t secs.

First, let's find how far it travels in t seconds:
dx= Vavg*dt
Vavg = (Vo +Vf)/2 = 0.5*Vf = 0.5 (a*t) = 5t m/s
dx= Vavg*dt = 5t m/s * t s = 5t^2 m

Second, let's find how far it travels in 3t seconds:
dx= Vavg*dt
Vavg = (Vo +Vf)/2 = 0.5*Vf = 0.5 (a*3t) = 15t m/s
dx= Vavg*dt = 15t m/s * 3t s = 45t^2 m

In terms of displacement in t seconds, displacement in 3ts is (45t^2)/(5t^2) = X9 greater

A much simpler way to see that it's 9 not 3 is to think about it conceptually. If t triples and x triples too, this means the two are directly proportionally?? This necessarily implies constant velocity!! If u do 60mi/hr on the highway, by definition you move 1mi/min however when u accelerate, v increases so by definition you cover greater distances for each unit of time.

Generally, it's safer to stick with the kinematic equations bcos they are more "foolproof" n can prevent errors like these, but ur intuition is right.

 

[vickpick]

Your intuition and reasoning are correct. If acceleration is constant, dx= Vavg*dt. Both formulae are correct, you can derive
dx=xo + vot +0.5 (at^2) with calculus. I think your mistake stems from ur calculation of Vavg:

I think the problem asks by what factor does the object travel in 3t secs compared to t secs.

First, let's find how far it travels in t seconds:
dx= Vavg*dt
Vavg = (Vo +Vf)/2 = 0.5*Vf = 0.5 (a*t) = 5t m/s
dx= Vavg*dt = 5t m/s * t s = 5t^2 m

Second, let's find how far it travels in 3t seconds:
dx= Vavg*dt
Vavg = (Vo +Vf)/2 = 0.5*Vf = 0.5 (a*3t) = 15t m/s
dx= Vavg*dt = 15t m/s * 3t s = 45t^2 m

In terms of displacement in t seconds, displacement in 3ts is (45t^2)/(5t^2) = X9 greater

A much simpler way to see that it's 9 not 3 is to think about it conceptually. If t triples and x triples too, this means the two are directly proportionally?? This necessarily implies constant velocity!! If u do 60mi/hr on the highway, by definition you move 1mi/min however when u accelerate, v increases so by definition you cover greater distances for each unit of time.

Generally, it's safer to stick with the kinematic equations bcos they are more "foolproof" n can prevent errors like these, but ur intuition is right.

right on 👍

 

[AG22]

Thanks for your explanation 🙂

 

[0Complications]

On free fall questions, I think the EK method is the easiest.

Say you start from zero and fall for 6 seconds, you can easily calculate the distance by figuring out the average velocity:

6s x10 m/s = 60m/s2

Now just divide your add V0 + Vf and divide by 2

V0 = 0

So it's 60/2 = 30m/s is avg velocity.

Now just multiply that by your time:

30m/s x 6 sec = 180 m


I don't like to use the kinematics unless I absolutely have to, so these types of shortcuts can save a lot of time and make the calculations more intuitive. At least for me.