Calculating Speed and Distance Using Acceleration and Time in Mechanics

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A car accelerates from rest at 3.00 m/s² and travels 20.0 m in 1.40 seconds, prompting questions about calculating speed at two points, distance to the first point, and time taken to reach it. Participants emphasize using kinematic equations, specifically noting that instantaneous acceleration is not necessary for this problem. The discussion suggests starting with the equations x = x0 + Vt + 0.5at² and Vf² - Vi² = 2ax to derive the needed values. Confusion arises among participants about identifying the correct variables and equations, particularly regarding initial speed and acceleration. Understanding these concepts is crucial for solving the problem effectively.
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Homework Statement


A car accelerates horizontally from rest on a level road at a constant acceleration of 3.00 m/s^2. Down the road, it passes through two photocells ("electric eyes" designated by 1 for the first one and 2 for the second one) that are separated by 20.0m. The time interval to travel this 20.0m distance as measured by the electric eyes is 1.40s.

a. Calculate the speed of the car as it passes each electric eye.
b. How far is it from the start to the first electric eye?
c. How long did it take the car to get to the first electric eye?


I know that this equation deals with Instantaneous acceleration or Instantaneous velocity, I just don't know where to start and how to get the ball rolling on working out the solution.
 
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I have the same problem for homework. But I'm not even sure where to start. That link doens't even help me because it says the same thing in my book. Which equations do I need to start with?
 
I always start problems like this with a good drawing, it always makes things easier. To solve problems like this you only need the two golden equations*. You don't need to consider any instantaneous acceleration or velocity in this problem(because there isn't any). I would start by solving question a. I think either one of the equations will actually work for part a.



*HINT: x = x0 + Vt+.5at^2 and Vf^2-Vi^2=2ax.
 
Thanks! I drew a picture but it just wasn't working.

Alright so let me get this straight:

a= 3.00 m/s2
x= 20.0 m
t= 1.40 s
v= 20.0 m/1.4 s

...which would mean that I have to solve for x0?

Sorry, but I have never taken a physics class until now and I'm just so confused.
 
You can't work out speed like that since the car is accelerating. For the speed at the first gate you have a distance a time and you know the acceleration. Now what you need to find is the initial speed. What equation has these elements in it so you can find initial speed?
 
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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