Answer:
a) The taxi is 107 s in motion
b) The average velocity is 26.2 m/s
Explanation:
First, the car travels with an acceleration of 2.00 m/s². The equations for position and velocity that apply for the car are:
x = x0 + v0 t + 1/2 a t²
v = v0 + a t
where
x = position at time t
x0 = initial position
v0 = initial speed
t = time
a = acceleration
v = speed
Let´s calculate how much distance and for how long the taxi travels until it reaches a speed of 29.0 m/s:
Using the equation for velocity:
v = v0 + a t
v - v0 / a = t
(29.0 m/s - 0 m/s) / 2 m/s² = t
<u>t = 14.5 s</u>
Then, in the equation for position:
x = x0 + v0 t + 1/2 a t²
x = 0 + 0 + 1/2 * 2.00 m/s² * (14.5 s)²
<u>x = 210 m</u>
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Then, the vehicle travels at constant speed for 87 s. The distance traveled will be:
x = v * t
x = 29.0 m/s * 87.0 s = <u>2.52 x 10³ m</u>
Lastly the car stops (v = 0) in 5 s. In this case, the car has a constant negative acceleration:
Using the equation for velocity:
v = v0 + a t
if v=0 in 5 s, then:
0 = 29.0 m/s + a * 5.00 s
a = -29.0 m/s / 5.00 s
a = -5.80 m/s²
Using now the equation for the position, we can calculate how far has the taxi traveled until it came to stop:
x = x0 + v0 t + 1/2 a t²
x = 0 + 29.0 m/s * 5.00 s -1/2 * 5.80 m/s² * (5.00s)²
<u>x = 72.5 m</u>
a) The taxi has been in motion for:
Total time = 14.5 s + 87.0 s + 5.00s = <u>107 s</u>
Note that we have always used x0 = 0, then, we have calculated the displacement for each part of the trip.
Adding all the displacements, we will get the total displacement:
Total displacement = 210 m + 2.52 x 10³ m + 72.5 m = 2.80 x 10³ m
Average speed = total displacement / total time
<u>Average speed = 2.80 x 10³ m / 107 s = 26.2 m/s </u>
