A car slows down uniformly from a speed of 29.0 m/s to rest in 8.00s. How far does it travel in that time.
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I know that the equation is X = X0 + V0(t) + .5(at^2)
I cant remeber where to put what numbers though.
Thanks for the help!
To find the distance the car travels in the given time, you can use the equation X = X0 + V0t + 0.5at^2, where:
X is the final position or distance traveled
X0 is the initial position (which we assume to be 0 since the car comes to rest)
V0 is the initial velocity of the car
t is the time taken
a is the acceleration (which we assume to be constant, since the car slows down uniformly)
So, let's substitute the known values into the equation and solve for X:
X = X0 + V0t + 0.5at^2
Since the car comes to rest, its final velocity V is 0 m/s. Therefore:
0 = V0 + at
Rearranging this equation to solve for a:
a = -V0/t
Substituting the given values:
a = -29.0 m/s / 8.00 s
a = -3.63 m/s^2
Now we have the acceleration, we can substitute it back into the equation:
X = X0 + V0t + 0.5at^2
X = 0 + 29.0 m/s * 8.00 s + 0.5 * ( -3.63 m/s^2) * (8.00 s)^2
Simplifying this equation will give you the distance traveled by the car in that time.
uniform acceleration means that the average speed is
... (initial + final) / 2 ... (29.0 + 0.0) / 2
distance is ... (average speed) * (time)