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A vehicle starts from rest with a uniform acceleration until a velocity of 48 km/h is reached. It then moves with a constant speed until it decelerates and finally stops 600 m from the starting point. The whole journey takes 1 minute, and the magnitude of the deceleration is twice that of the acceleration. Calculate:

The time (in seconds) it takes the vehicle to reach a velocity of 48 km/h.
The time (in seconds) the vehicle moves with a constant speed.
The time (in seconds) it takes the vehicle to decelerate and stop.
The acceleration of the vehicle (give your answer in m/s2).
The distance the vehicle travels with constant velocity (give your answer in m)/

Respuesta :

onyebuchinnaji

The time it takes the vehicle to reach a velocity of 48 km/h, t₁ = 20 s

The time the vehicle moves with a constant speed, t₂ = 30 s

The time it takes the vehicle to decelerate and stop, t₃ = 10 s

The acceleration of the vehicle is 0.66 m/s².

The distance the vehicle travels with constant velocity is 400 m.

The given parameters:

  • Velocity of the vehicle, u = 48 km/h = 13.33 m/s
  • Distance traveled by the vehicle, d = 600 m
  • Total time of the Journey, t = 1 minute = 60 s
  • Deceleration = 2a

The total distance traveled by the vehicle is calculated as follows;

[tex]s_t = s_1 + s_2 + s_3\\\\s_t= ( \frac{1}{2} bh ) + (bh)+( \frac{1}{2} bh)\\\\s_t = (\frac{1}{2} t_1v_1)+ (t_2v_2)+ (\frac{1}{2} t_3v_3)\\\\600 = (0.5t_1\times 13.33) + (t_2\times 13.33) + (0.5 \times t_3 \times 13.33)\\\\600 = 6.67t_1 + 13.33t_2 + 6.67t_3 \\\\600 = 6.67(t_1 + t_3) + 13.33t_2 \ --- (1) \\\\[/tex]

[tex]t_1 + t_2 + t_3 = 60\\\\t_1 + t_3 = 60 -t_2[/tex]

[tex]600 = 6.67(60-t_2) + 13.33t_2\\\\600 = 400.2 - 6.67t_2 + 13.33t_2\\\\600 = 400.2 + 6.67t_2\\\\199.8 = 6.67t_2\\\\t_2 = \frac{199.8}{6.67} \\\\t_2 = 29.96 \ s[/tex]

[tex]t_1 = 60-t_2 -t_3\\\\t_1 = 60 -29.96 - t_3\\\\t_1 = 30.04 -t_3[/tex]

The acceleration of the vehicle;

[tex]a = \frac{v-u}{t_1} \\\\a = \frac{13.33-0}{t_1} \\\\a = \frac{13.33}{t_1}\\\\t_1 = \frac{13.33}{a}[/tex]

The deceleration of the vehicle;

[tex]dec =\frac{-13.33}{t_3} \\\\|dec| = 2a\\\\\frac{13.33}{t_3} = \frac{2 (13.33)}{t_1} \\\\13.33t_1 = 2(13.33)t_3\\\\t_1 =2t_3[/tex]

[tex]t_1 = 30.04 - t_3 \\\\2t_3 = 30.04 -t_3\\\\3t_3 = 30.04\\\\t_3 = \frac{30.04}{3} \\\\t_3 = 10.01 \ s[/tex]

[tex]t_1 = 2t_3\\\\t_1 = 2(10.01)\\\\t_1 = 20.02 \ s[/tex]

Thus, we can conclude the following:

  • The time it takes the vehicle to reach a velocity of 48 km/h, t₁ = 20 s
  • The time the vehicle moves with a constant speed, t₂ = 30 s
  • The time it takes the vehicle to decelerate and stop, t₃ = 10 s

The acceleration of the vehicle is calculated as follows;

[tex]a = \frac{13.33}{t_1} \\\\a = \frac{13.33}{20} \\\\a = 0.66 \ m/s^2[/tex]

The distance the vehicle travels with constant velocity is calculated as;

[tex]s_2 = v_2 t_2\\\\s_2 = 13.33 \times 30\\\\s_2 = 400 \ m[/tex]

Learn more about velocity time graph here: https://brainly.com/question/14086970

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