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 leav

Question

dimaraw

1 month ago

13

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 leav

es the crossing, how much time is required until its speed reaches 32 m/s?

Physics

2 answers:

kicyunya [3.2K] · Answer 1 · 2026-02-28T09:43:00+03:00

Response:

Once it has crossed, the locomotive requires 17.6 seconds to achieve a speed of 32 m/s.

Details:

The locomotive's acceleration is 1.6 $m/s^2$

The duration taken to pass the crossing is 2.4 seconds.

We can apply the motion equation, v = u + at, where v represents final velocity, u indicates initial velocity, a denotes acceleration, and t signifies time.

When the speed reaches 32 m/s, we have v = 32 m/s, u = 0 m/s, and a= 1.6 $m/s^2$ .

32 = 0 + 1.6 * t

t = 20 seconds.

Therefore, the locomotive attains a speed of 32 m/s after 20 seconds, and it passes the crossing in 2.4 seconds.

Thus, after clearing the crossing, it takes an additional 17.6 seconds to reach the speed of 32 m/s.

Keith_Richards [3.2K] · Answer 2 · 2026-02-28T09:41:27+03:00

After the train has passed the crossing, it takes around 14 seconds for its speed to accelerate to 32 m/s.

Additional explanation

Acceleration refers to how quickly velocity changes.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

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

u = initial velocity (m / s)

t = time taken (s)

d = distance (m)

Let’s proceed to solve the problem!

Given:

a = 1.6 m/s²

d = 20.0 m

t₁ = 2.4 s

v₂ = 32 m/s

Find:

Δt =?

Solution:

First, we will determine the initial velocity of the locomotive as it enters the crossing.

$d = ut + \frac{1}{2}at_1^2$

$20 = u(2.4) + \frac{1}{2}(1.6)(2.4)^2$

$20 = u(2.4) + 4.608$

$u(2.4) = 20 - 4.608$

$u = 15.392 \div 2.4$

$u = (481 \div 75) ~ m/s$

Next, we will compute the total duration required for the train to reach 32 m/s.

$v_2 = u + at_2$

$32 = (481 \div 75) + 1.6t_2$

$t_2 = (32 - \frac{481}{75}) \div 1.6$

$t_2 \approx 16 ~ seconds$

Ultimately, the time it takes for the locomotive to attain a speed of 32 m/s after leaving the crossing is:

$\Delta t = t_2 - t_1$

$\Delta t = 16 - 2.4$

$\Delta t \approx 14 ~ seconds$

Learn more

Velocity of Runner:

Kinetic Energy:

Acceleration:

The Speed of Car:

Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

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