Velocity/displacement 11th grade physics

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Halving the initial velocity of a skidding truck significantly reduces its total displacement. The reasoning is that a lower velocity results in less distance covered over the same time period, leading to a displacement that could be around one-third of the original. The discussion emphasizes using mathematical formulas to derive solutions rather than relying on trial and error methods, which are deemed ineffective long-term. The distinction between displacement and distance is clarified, noting that displacement accounts for direction while distance does not. Overall, understanding the relationship between velocity, acceleration, and displacement is crucial in physics.
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Homework Statement


A truck that is skidding out of control has its initial velocity halved. How will this effect the total displacement of the skidding truck, vs the amount of displacement if the truck were at its original velocity


Homework Equations


displacement= Vi(t)+1/2AT^2 Vi=initial velocity... A=acceleration...T=time


The Attempt at a Solution


I believe that the result would be around less than half the displacement. My reasoning is that the total time to stop would be less, because there is less of a velocity, and also that since the velocity is lower, that the truck would be covering less distance throughout the whole time... maybe 1/3 or around that as a ballpark guess.
 
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Try putting some variables in the displacement function, for example Vi = a m/s, A = a/4 m/s^2. Now put Vi = 2a. Compare the expressions then.
 
kbaumen said:
Try putting some variables in the displacement function, for example Vi = a m/s, A = a/4 m/s^2. Now put Vi = 2a. Compare the expressions then.
dear frnd... kbaumen never encourage your colleagues to put values and solve questions...

trial and error methods are very uncivilized and baseless methods..

they may solve a problem for you but they shall cease to help you in the long run...

use maths to help you always...

v=u+at
a=v/t
v/2=u+a't
a'=v/2t

s1=1/2 v/t *t^2
s2=1/2 v/2t *t^2

s1/s2=2vt/vt
s1/s2=2
s1=2s2

assumptions=the truck starts from rest ..the acceleration is constant.. and the displacements are measured in the same time interval.
 
physixguru said:
dear frnd... kbaumen never encourage your colleagues to put values and solve questions...

trial and error methods are very uncivilized and baseless methods..

they may solve a problem for you but they shall cease to help you in the long run...

use maths to help you always...

Thanks for the advice.
 
The formula for displacement that you used is the same formula i use for solving for distance? why is that? this is to lax113
 
displacement is distance, but it takes into account direction, whereas distance doesn't...so its the same formula for both
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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