Simple motorboat kinematics problem

AI Thread Summary
A student is struggling with a motorboat kinematics problem involving uniform deceleration from 75 km/h to 40 km/h over 50 meters. They express frustration with memorizing formulas and seek a deeper understanding through calculus. The discussion outlines the derivation of kinematic equations, emphasizing the integration of acceleration to find velocity and position functions. A key insight is provided on eliminating time from the equations, which helps solve the problem. The student appreciates the clarification on using the relationship between velocity and acceleration to derive the necessary formula.
maxbashi
Messages
18
Reaction score
0
I'm stuck doing this problem in my first year mechanics class. I feel sort of dumb for not being able to get it. The teacher wants us to just memorize the formulas but I really really really don't want to do that - I'm trying to just do it using the basics and calculus. I can obviously do this problem by plugging in numbers, but I'm looking for a way to do it from scratch.


Homework Statement


A motorboat traveling on a straight course slows dow uniformly from 75 km/h to 40 km/h in a distance of 50 m. What is the magnitude of its acceleration?


Homework Equations


The one that you're "supposed" to used to get the right answer is v^2=vo^2+2a(x-xo).
I'm just trying to figure out why one would derive it this way - why the square speed? Obviously it's dimensionally needed to get m/s^2 for the answer... but I'm just looking for any explanation I can get.

The Attempt at a Solution


several approaches I've tried...
v(t)=(75-at)km/h - but don't know how to find what t would be, since only the distance (50m) is given

average acceleration = (75-40)/t... again, the problem is that I can't find t.

I would appreciate any help you guys could give me. And if I can provide any more information about my thoughts so far let me know. Thanks
 
Physics news on Phys.org
From scratch eh? Okay:

You start with the acceleration as a function of time:

a(t) = a = const. (1)​

Integrate this once with respect to time to get the velocity as a function of time :

v(t) = v0 + at (2)​

Here, v0 is the constant of integration, and it is obviously equal to the initial velocity (which you'll see if you set t = 0). Integrate this function with respect to time to get the position function:

x(t) = x0 + v0t + (1/2)at2 (3)​

Now you've derived all but one of the kinematics formulae. The last one makes no explicit mention of time. To eliminate t from the equations, solve for t in (2):

t = (v - v0) / a​

Substitute this into (3), rearrange, and you will end up with the result that:

v2 - v02 = 2a(x - x0) (4)​

An easier way to derive (4) is to use the work-energy theorem. As soon as your prof teaches you what that is, try it!
 
Thank you so much! Eliminating t by setting it equal to (v-vo)/a was just what I was missing. Thanks!
 
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}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top