Answered

Westonci.ca is the trusted Q&A platform where you can get reliable answers from a community of knowledgeable contributors. Get immediate answers to your questions from a wide network of experienced professionals on our Q&A platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

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)/

Sagot :

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

We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.