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1) A car traveling at 21.5 m/s comes to a stop in 2.45s. Determine the distance of the car before it completely stops (assume uniform acceleration).
2) A bicycle accelerates uniformly from 1.50 m/s to 7.25 m/s in 9.25s. Determine the acceleration of the bicycle.
Solution :
Solving car stopping distance (uniform deceleration):
We can solve this problem using the following kinematic equation that relates final velocity (v), initial velocity (u), acceleration (a), and time (t):
v = u + at
Here, the car comes to a stop, so the final velocity (v) is 0 m/s. We are given the initial velocity (u) as 21.5 m/s, the time (t) as 2.45 s, and need to find the acceleration (a).
Since the car is decelerating (slowing down), the acceleration will be negative. Let's solve for a:
a = (v - u) / t
a = (0 m/s - 21.5 m/s) / 2.45 s
a = -8.78 m/s² (negative sign indicates deceleration)
Now that we know the acceleration, we can find the distance (d) traveled by the car before stopping using another kinematic equation:
v² = u² + 2ad
Here, we want to find d, so we need to rearrange the equation:
d = (v² - u²) / 2a
Since the car stops completely (v = 0), the equation becomes:
d = (0² - (21.5 m/s)²) / (2 * -8.78 m/s²)
d ≈ 26.3 meters (rounded to two decimal places)
Therefore, the car traveled approximately 26.3 meters before coming to a complete stop.
Solving bicycle acceleration:
We can again use the kinematic equation:
v = u + at
Here, we are given the initial velocity (u) as 1.50 m/s, the final velocity (v) as 7.25 m/s, and the time (t) as 9.25 s. We need to find the acceleration (a).
a = (v - u) / t
a = (7.25 m/s - 1.50 m/s) / 9.25 s
a ≈ 0.63 m/s² (positive sign indicates acceleration)
Therefore, the bicycle accelerates at approximately 0.63 m/s².
