acar traveling along a straight roed increases it's speed from 30m/s to 50m/s a distance of 300m .If the acceleration is constant how much time elapses while the car moves this distance?
constant acceleration means speed is linear with time
v = 30 + a t
that means you can use the average speed, 40 m/s
300 m / 40 m/s = 7.5 seconds
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proof it works:
50 = 30 + a t so a t = 20 and t = 20/a
300 = 30 t + .5 a t^2
300 = 30 (20/a) + .5(20)(20)/a
300 = 40 (20/a)
300 a = 800
a = 8/3
then t = 20 * 3/8 = 7.5 seconds sure enough
To determine the time it takes for the car to travel a distance with a constant acceleration, we can use the equation:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
In this case, the initial velocity (u) is 30 m/s, the final velocity (v) is 50 m/s, and the distance traveled (s) is 300 m. We need to find the time (t).
We can rearrange the equation to solve for time:
t = (v - u) / a
Substituting the given values:
t = (50 m/s - 30 m/s) / a
To find the acceleration (a), we can use another equation:
v^2 = u^2 + 2as
Where:
v = final velocity
u = initial velocity
a = acceleration
s = distance
Rearranging this equation to solve for acceleration:
a = (v^2 - u^2) / (2s)
Substituting the given values:
a = (50 m/s)^2 - (30 m/s)^2 / (2 * 300 m)
Now we have the value of acceleration (a), which we can substitute back into the equation to find the time (t):
t = (50 m/s - 30 m/s) / [(50 m/s)^2 - (30 m/s)^2 / (2 * 300 m)]
Evaluating this expression will give you the time elapsed while the car travels the distance of 300m.
Use the kinematic formula vf^2=vi^2+2ad.
50^2=30^2+2*a*300
Solving for "a" gives 2.67 m/s/s, or 3 m/s/s if you must use significant digits.
Remember that when you have 3 knowns and acceleration is constant, you can use kinematics to find an unknown.