Solve Acceleration Problem: Car Overtaken After 4 sec.

  • Thread starter Thread starter bond.173
  • Start date Start date
AI Thread Summary
The problem involves two cars: one entering the speedway with an acceleration of 6.8 m/s² and the other traveling at a constant speed of 62 m/s. After 4 seconds of acceleration, the entering car reaches a speed of 27.2 m/s. The position equations for both cars are established, with the entering car's equation incorporating its acceleration and the other car's equation reflecting its constant speed. By setting the position equations equal, the time required for the entering car to catch up to the other car is calculated to be approximately 10.2 seconds. This analysis effectively demonstrates the use of kinematic equations to solve for time in acceleration scenarios.
bond.173
Messages
1
Reaction score
0
After refueling, car has an acceleration whose magnitude is 6.8 m/s^2, after 4 sec. he enters the speedway. At the same instant, another car and traveling at a constant speed of 62 m/s overtakes and passes the entering car. If the entering car maintains its acceleration, how much time is required for it to catch the other car?
 
Physics news on Phys.org
You will want to set up separate position equations for each car. Call the point at the end of the runway from the maintenance pit, where the entering car returns to the race track, x = 0 ; also call the time when the entering car returns, and the moving car is just passing it, t = 0.

At that moment, the car in the race is moving at a constant 62 m/sec. What would its position kinematic equation be (make it, say, x_B = ... ) ? The entering car has been accelerating at 6.8 m/(sec^2) for 4 seconds by t = 0 when it reaches x = 0. What would this car's position equation be ( x_A = ... ) ?

When the entering car catches up to the car already in the race, you will have x_A = x _B . On setting your two equations equal, you will have something you can rearranging into a single quadratic equation you can solve for t . There will probably be two solutions, but since the equation only applies for t = 0 and afterwards, you can throw away any negative answer for t.
 
Car 1's velocity upon entering the track:
\ a = \frac{v - v_0}{t}
\ v = a·t + v_0 = (6.8\:m/s^{2})(4\:s) + 0 = 27.2\:m/s

Car 1's position on track at time t:
\ d = vt + \frac{1}{2}at^{2}

Car 2's position on track at time t:
\ d = vt

Set these positions equal to one another and solve for t:
\ v(car 1)t + \frac{1}{2}at^{2} = v(car 2)t
\ v(car 1)t + \frac{1}{2}at^{2} - v(car 2)t = 0
\ t[v(car 1) - v(car2) + \frac{1}{2}at] = 0

t = 0 is the first solution that jumps out at you, but it's uninteresting, because it just says the car 2 overtook car 1 at the beginning, which we already know from the problem statement. So you want to t for the bracketed part:
\ v(car 1) - v(car2) + \frac{1}{2}at = 0
\ \frac{1}{2}at = v(car2) - v(car1)
\ t = 2\frac{v(car2) - v(car1)}{a} = 2\frac{62\:m/s - 27.2\:m/s}{6.8\:m/s^{2}} = 10.2\:s
 
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}...
Back
Top