Minimum deaceleration untll the car stops

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To solve the problem of a car stopping on a 35m-long shoulder from a speed of 25 mi/h, the correct approach involves converting the speed to meters per second. The equation a = v^2 / (2x) is appropriate for calculating the required minimum acceleration. The initial calculations yielded incorrect results due to improper unit conversion. After correcting the conversion from miles per hour to meters per second, the values for acceleration and time can be accurately determined. Proper unit conversion is crucial for solving physics problems effectively.
forensicchemgirl
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what am i doing wrong??
every time i try to answer this i get t=.2s and a=-1651ms^-2

the problem reads...
A car traveling at 25mi/h is to stop on a 35m-long shoulder of the road. (a) What is the required magnitude of he minimum acceleration? (b)How much time will elapse during this minimum deaceleration untll the car stops.

i used the equation...
a=v^2/2x



 
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i think the first time is to convert 25mi/h into meters per second.

~Amy
 
Yes, use 1 mile =1.6km, and don't forget to convert to m/s, so you need to change out of km/h. 1h=3600s. That equation is right.
 
yes i did convert to m/s wrong thanks for the help.:smile: :-p
 
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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