Question

Physics Help Needed?

Answer 1

Total distance`=783.dot3m`
Averge speed `approx 16.2m//s`

Explanation:

Three steps are involved in the running of train.

1. Starts from rest from say station 1 and is accelerates for `10 s`.
   Distance `s_1` traveled in these 10 s.
   `s_1=ut+1/2at^2`
   Since it starts from rest, therefore, `u=0`
   `:. s_1=1/2xx2xx10^2`
   `s_1=100m`

2. Runs for next `30 s` at constant speed.
   Distance run `s_2=speed xx time` .....(1)
   Speed at the end of acceleration `v=u+at`
   `v=2xx10=20m//s`. Inserting value of `v` in (1), we obtain
   `s_2=20xx30=600m`

3. Decelerates till it stops, i.e., from speed of `20 m//s` to zero.
   Using the expression
   `v=u+at`
   we find time `t_3`taken to come to stop.
   `0=20-2.4xxt_3`
   `=>t_3=20/2.4 =8.dot3s`
   Also use
   `v^2-u^2=2as`
   to find out distance `s_3` traveled in this time `t_3`

`0^2-20^2=2xx-2.4xxs_3`
`=>s_3=400/4.8=83.dot3m`
Total distance traveled by the train `=s_1+s_2+s_3`
`=100+600+83.dot3=783.dot3m`

Average speed`="Total distance traveled"/"Total time taken"`
`=(783.dot3)/(10+30+8.dot3)`
`approx 16.2m//s`

Answer 2

Here's what I got.

Explanation:

One interesting thing to notice here is that the subway's acceleration and deceleration are not equal.

This should tell you that it takes less time for the subway to come to a complete stop from its max speed than it takes it to reach max speed.

Implicitly, this should also tell you that the subway accelerates over a longer distance than the distance it needs to come to a complete stop.

So, your goal here is to find two things

• the total displacement of the subway, i.e. how far it is from its starting point when it stops
• the total time needed to get from its starting point to its destination

Since the subway is traveling in a straight line, you can use distance instead of displacement and speed instead of velocity.

Break the motion of the subway in three stages

• From rest to maximum speed

The subway starts from rest and moves with an acceleration of `"2.0 m s"^(-2)` for a total time of `"10 s"`. Think about what acceleration means.

An acceleration of `"2.0 m s"^(-2)` tells you that with every passing second, the speed of the subway increases by `"2.0 m s"^(-1)`. You an describe its final speed in terms of its initial speed, `v_0`, its acceleration, `a`, and the time of motion, `t`, using the equation

`color(blue)(v_f = v_0 + a * t)`

Well, if it starts from rest and moves for `"10 s"`, it follows that its maximum speed will be

`v_"max" = overbrace(v_0)^(color(purple)(=0)) + "2.0 m s"^color(red)(cancel(color(black)(-2))) * 10color(red)(cancel(color(black)("s"))) = "20 ms "^(-1)`

The distance traveled for this first stage will be equal to

`color(blue)(d = overbrace(v_0 * t)^(color(purple)(+0)) + 1/2 * a * t^2)`

`d_1 = 1/2 * "2.0 m" color(red)(cancel(color(black)("s"^(-2)))) * (10^2) color(red)(cancel(color(black)("s"^2))) = "100 m"`

• Moving at constant speed

Once the subway's peed reaches `"20 m s"^(-1)`, it stops accelerating and begins to move at constant speed.

A peed of `"20 m s"^(-1)` tells you that with every passing second, the subway travels a distance of `"20 m"`. This means that you have

`color(blue)(d = v * t)`

`d_2 = "20 m" color(red)(cancel(color(black)("s"^(-1)))) * 30color(red)(cancel(color(black)("s"))) = "600 m"`

• From maximum speed to rest

This time, the subway starts from maximum speed and must come to a complete stop. You can determine the distance it takes for it do so by using the equation

`color(blue)(v_s^2 = v_0^2 + 2 * a * d_3)" "`, where

`v_s` - its final speed
`v_0` - its speed at the moment it begins to decelerate, here equal to `v_'max"`
`d_3` - its stopping distance

Now, it is very important to understand that you must use

`a = -"2.4 m s"^(-2)`

The subway is moving west, as indicated by the symbol `["W"]`. In order for it to stop, deceleration must be oriented in the opposite direction, i.e. east, `["E"]`.

If you take west to be the positive direction, you must take east to be the negative one.

So, the stopping distance will be

`overbrace(v_s)^(color(purple)(=0)) = v_"max"^2 - 2 * "2.4 m s"^(-2) * S`

`d_3 = v_"max"^2/(2 * "2.4 m s"^(-2))`

`d_3 = (20^2 "m"^color(red)(cancel(color(black)(2))) * color(red)(cancel(color(black)("s"^(-2)))))/(2 * 2.4 color(red)(cancel(color(black)("m")))color(red)(cancel(color(black)("s"^(-2))))) = "83.33 m"`

Notice that, as predicted, the deceleration distance is indeed shorter than the acceleration distance.

The time it takes for the subway to decelerate will be

`overbrace(v_f)^(color(purple)(=0)) = v_"max" - "2.4 m s"^(-2) * t_d`

`t_d = (20color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1)))))/(2.4color(red)(cancel(color(black)("m")))"s"^color(red)(cancel(color(black)(-2)))) = "8.33 s"`

The total distance covered by the subway is

`d_"total" = d_1 + d_2 + d_3`

`d_"total" = "100 m" + "600 m" + "83.33 m" = "783.33 m"`

The total time needed to cover this distance

`t_"total" = "10 s" + "30 s" = "8.33 s" = "48.33 s"`

The average speed of the subway was - remember that I'm using distance instead of displacement!

`color(blue)("avg. speed" = "the distance you traveled"/"how long it took you to do it")`

`bar(v) = "783.33 m"/"48.33 s" = color(green)("16.2 ms"^(-1))`

I'll leave the answer rounded to three sig figs.