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A locomotive is accelerating at 1.6 m/s2. it passes through a 20.0-m-wide crossing in a time of 2.4 s. after the locomotive leaves the crossing, how much time is required until its speed reaches 32 m/s?

Respuesta :

johanrusli

It required about 14 seconds after the locomotive leaves the crossing until its speed reaches 32 m/s.

Further explanation

Acceleration is rate of change of velocity.

[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]

[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]

a = acceleration (m / s²)v = final velocity (m / s)

u = initial velocity (m / s)

t = time taken (s)

d = distance (m)

Let us now tackle the problem!

Given:

a = 1.6 m/s²

d = 20.0 m

t₁ = 2.4 s

v₂ = 32 m/s

Unknown:

Δt = ?

Solution:

Initially, we calculate the initial speed of the locomotive when entering the crossing.

[tex]d = ut + \frac{1}{2}at_1^2[/tex]

[tex]20 = u(2.4) + \frac{1}{2}(1.6)(2.4)^2[/tex]

[tex]20 = u(2.4) + 4.608[/tex]

[tex]u(2.4) = 20 - 4.608[/tex]

[tex]u = 15.392 \div 2.4[/tex]

[tex]u = (481 \div 75) ~ m/s[/tex]

Next we calculate the total time needed for the locomotive to reach 32 m/s

[tex]v_2 = u + at_2[/tex]

[tex]32 = (481 \div 75) + 1.6t_2[/tex]

[tex]t_2 = (32 - \frac{481}{75}) \div 1.6[/tex]

[tex]t_2 \approx 16 ~ seconds[/tex]

Finally, the time needed for the locomotive to reach a speed of 32 m/s after leaving the crossing is:

[tex]\Delta t = t_2 - t_1[/tex]

[tex]\Delta t = 16 - 2.4[/tex]

[tex]\Delta t \approx 14 ~ seconds[/tex]

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Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate , Sperm , Whale , Travel

Ver imagen johanrusli
onyebuchinnaji

The time required to reach the speed of 32 m/s is 16 s.

The given parameters;

acceleration, a = 1.6 m/s²

distance traveled, d = 20 m

time of motion, t = 2.4 s

final velocity, v = 32

The initial velocity is calculated as follows;

[tex]s = ut + \frac{1}{2} at^2\\\\20 = u(2.4) + 0.5 \times 1.6 \times (2.4)^2\\\\20 = 2.4u + 4.608 \\\\2.4u = 20 - 4.608\\\\2.4u = 15.39\\\\u = \frac{15.39}{2.4} = 6.41 \ m/s[/tex]

The time required to reach the speed of 32 m/s;

[tex]v= u+ at\\\\32 = 6.41 + 1.6t\\\\1.6t = 32-6.41 \\\\1.6t = 25.59\\\\t = \frac{25.59}{1.6} \\\\t = 15.99 \ s\approx 16 \ s[/tex]

Thus, the time required to reach the speed of 32 m/s is 16 s.

Learn more here: https://brainly.com/question/2668062

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