Question

Поезд двигаясь под уклон прошел за 20 сек путь 340 м и развил скорость 19 м/с С каким ускорением двигался поезди какой была скорость в начале уклона

Answer 1

 

по формуле s=(v0 + v)t /2  найдем v0

340=(v0 + 19)20 /2=(v0 + 19)*10

разделим обе части на 10 и получим,что

34=v0 + 19 отсюда v0=34-19=15 м/с

найдем ускорение по формуле

a=(v-v0)/t=4/20=0,2 м/с^2

Answer 2

Calculating Train's Acceleration and Initial Speed on an Incline

To calculate the train's acceleration and initial speed on the incline, we can use the given information about the time, distance, and final speed of the train.

Given: - Time taken to travel the distance on the incline, *t* = 20 seconds - Distance traveled on the incline, *d* = 340 m - Final speed of the train, *v* = 19 m/s

Calculating Acceleration:

To find the acceleration of the train, we can use the following kinematic equation: \[ d = v_i t + \frac{1}{2} a t^2 \] where: - *d* = distance traveled - *v_i* = initial speed - *t* = time taken - *a* = acceleration

Rearranging the equation to solve for acceleration (*a*): \[ a = \frac{2(d - v_i t)}{t^2} \]

Substituting the given values: \[ a = \frac{2(340 - 19 \times 20)}{20^2} \]

\[ a = \frac{2(340 - 380)}{400} \]

\[ a = \frac{2 \times (-40)}{400} \]

\[ a = \frac{-80}{400} \]

\[ a = -0.2 \, m/s^2 \]

The acceleration of the train on the incline is -0.2 m/s^2.

Calculating Initial Speed:

To find the initial speed of the train on the incline, we can use the following equation of motion: \[ v = v_i + at \] where: - *v* = final speed - *v_i* = initial speed - *a* = acceleration - *t* = time taken

Rearranging the equation to solve for initial speed (*v_i*): \[ v_i = v - at \]

Substituting the given values: \[ v_i = 19 - (-0.2 \times 20) \]

\[ v_i = 19 + 4 \]

\[ v_i = 23 \, m/s \]

The initial speed of the train on the incline was 23 m/s. So, the train's acceleration on the incline is -0.2 m/s^2, and its initial speed on the incline was 23 m/s.