Car Velocity After Engine Cuts Out | Physics Problem

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A car traveling at +31.0 m/s experiences an engine failure, leading to a deceleration over two intervals: the first lasting ten seconds with average acceleration a1, and the second lasting five seconds with average acceleration a2. The final velocity after fifteen seconds is +24.5 m/s, and the ratio of the accelerations is a1/a2 = 1.67. The calculated velocity at the end of the first ten seconds is +26 m/s. The discussion emphasizes using initial and final velocities to determine average accelerations through simultaneous equations.
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Homework Statement


A car is traveling along a straight road at a velocity of +31.0 m/s when its engine cuts out. For the next ten seconds, the car slows down, and its average acceleration is a1[/size]. For the next five seconds, the car slows down further, and its average acceleration is a2[/size]. The velocity at the end of the fifteen-second period is +24.5 m/s. The ratio of the average acceleration values is a1[/size]/a2[/size] = 1.67. Find the velocity of the car at the end of the initial ten-second interval.

Homework Equations


VF = V0 + at

VF = Final velocity
V0 = Initial velocity
a = acceleration
t = time

The Attempt at a Solution



The answer is +26 m/s, but I don't know how to get it...

24.5 = 31.0 + a2(15)

a2 = -13/30 = -4.333...
a1 = -3.8534

VF = 31.0 + (-3.8534)(10)
VF = -7.534
 
Last edited:
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Ok, do you know that you can find the average acceleration just by doing simple operations with the inidial velocity and the final velocity?And with that you can find the values of a1 and a2 through use of simultaneous equations. You should be able to solve from here.
 
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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