Calculating Car Acceleration on a 1.2 Meter Hill in 2 Seconds

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To calculate the acceleration of a car traveling down a 1.2-meter hill in 2 seconds, use the equation d = (1/2)(a)(t)^2, where d is the distance, a is acceleration, and t is time. Since the initial velocity (vi) is zero, the equation simplifies to d = (1/2)(a)(t)^2. Rearranging this gives a = (2d) / (t^2). By substituting the values, you can find the acceleration without needing additional variables. This method effectively bypasses the need for force calculations.
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calculate the acceleration of a car that travels down a hillthat is 1.2 meters long in 2 seconds


a=f/m



all that i have down is the forula but there is no other variables.
--im stumped
 
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Use the constant acceleration motion equations. If you don't know them google motion equations
 
wow they are really confusing

is there a way i can get the solution without trying to figure them out?


how can i find the force if i don't even know the acceleration?
 
imm really pissed at my science teacher right now
 
wait... i have this nifty little equation you can use to calculate the acceleration with the final velocity... d=(vi)(t)+(1/2)(a)(t)^2 vi is the initial velocity, and that equals zero, so you can cut that out and use the d=(1/2)(a)(t)^2 and work the equation to equal a... that is all i am going to give you
 
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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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