A car, moving along a straight stretch of high-way, begins to accelerate at 0.0449 m/s2. It takes the car 64.6 s to cover 1 km.
How fast was the car going when it first began to accelerate?
Answer in units of m/s.
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d = Vo*t + 0.5*a*t^2 = 1,000 m.
64.6Vo + 0.5*0.0449*64.6^2 = 1,000.
Vo =
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To determine the initial velocity of the car when it began to accelerate, we can use the kinematic equation:
vf = vi + at
where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time. Rearranging the equation, we have:
vi = vf - at
Given that the car takes 64.6 s to cover 1 km (which is equal to 1000 m), we can convert the time to seconds and the distance to meters. Thus, t = 64.6 s and d = 1000 m.
Since the car starts from rest when it begins to accelerate, the final velocity (vf) is unknown. However, the final velocity can be calculated using the formula:
vf = (2 * d) / t
Substituting the given values, we have:
vf = (2 * 1000 m) / 64.6 s
= 31.03 m/s
Now we can substitute the values of vf, a, and t into the equation:
vi = vf - at
= 31.03 m/s - (0.0449 m/s^2) * 64.6 s
= 31.03 m/s - 2.903 m/s
= 28.13 m/s
Therefore, the car was initially going at a speed of 28.13 m/s when it first began to accelerate.