A taxi traveling along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 27.0 m/s. Then the vehicle travels for 41.0 s at constant speed until the brakes are applied, stopping the vehicle in a uniform manner in an additional 5.00 s. (a) How long is the taxi in motion (in s)? s (b) What is the average velocity of the taxi for the motion described? (Enter the magnitude in m/s.) m/s

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aristocles aristocles · Answer 1 · 2019-09-16T02:54:56+02:00

Answer:

Part a)

[tex]T = 59.5 s[/tex]

Part b)

v = 22.8 m/s

Explanation:

Part a)

Time taken by the taxi to reach the maximum speed is given as

[tex]v_f - v_i = at[/tex]

[tex]27 - 0 = 2(t)[/tex]

[tex]t = 13.5 s[/tex]

now it moves with constant speed for next 41 s and then finally comes to rest in next 5 s

so total time for which it will move is given as

[tex]T = 13.5 s + 41 s + 5 s[/tex]

[tex]T = 59.5 s[/tex]

Part b)

Distance covered by the taxi while it accelerate to its maximum speed is given as

[tex]d_1 = \frac{v_f + v_i}{2} t[/tex]

[tex]d_1 = \frac{27 m/s + 0}{2} (13.5)[/tex]

[tex]d_1 = 182.25[/tex]

Now it moves for constant speed for next 41 s so distance moved is given as

[tex]d_2 = (27)(41)[/tex]

[tex]d_2 = 1107 m[/tex]

Finally it comes to rest in next 5 s so distance moved in next 5 s

[tex]d_3 = \frac{v_f + v_i}{2} t[/tex]

[tex]d_3 = \frac{27 + 0}{2}(5)[/tex]

[tex]d_3 = 67.5 m[/tex]

so here we have total distance moved by it is given as

[tex]d = 182.25 + 1107 + 67.5 [/tex]

[tex]d = 1356.75 m[/tex]

average speed is given as

[tex]v = \frac{1356.75}{59.5}[/tex]

[tex]v = 22.8 m/s[/tex]