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Katyanochek1 [597]

3 years ago

12

A speed-time graph shows a car moving at 10 m/s for 10 s. The cars speed constantly decreases until it comes to a stop at 30 s.

Which describes the slope of the graph from 10 s to 30 s

2 answers:

egoroff_w [7]3 years ago

7 0

Actually the correct answer is going to be Linear, sloping downward.

Please mark Brainliest if this answer helped you.

Ray Of Light [21]3 years ago

5 0

speed of the car at t = 10 s is given as 10 m/s

now at t = 30 s car comes to rest so v = 0

now in this velocity time graph we can say slope will show the acceleration of the car as we know that acceleration is rate of change in velocity

$a = \frac{v_f - v_i}{\Delta t}$

as we know that

$v_f = 0$

$v_i = 10 m/s$

$\Delta t = (30 - 10) = 20 s$

now from above formula

$a = \frac{0 - 10}{20} = -0.5 m/s^2$

So slope of the graph will represent the acceleration of car which is 0.5 m/s^2

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2 years ago

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1 year ago

A light-rail train going from one station to the next on a straight section of track accelerates from rest at 1.1 m/s^2 for 20s.

svet-max [94.6K]

Answer:

A) The distance between the stations is 1430m

B) The time it takes the train to go between the stations is 80s

Explanation:

First we will calculate the distance covered for the first 20s.

From one the equations of kinematics for linear motion

$S = ut + \frac{1}{2}at^{2} \\$

Where $S$ is distance traveled

$u$ is the initial velocity

$t$ is time

and $a$ is acceleration

Since the train starts from rest, $u$ = 0 m/s

Hence, for the first 20s

$a =$ 1.1 m/s²; $t =$ 20s, $u$ = 0 m/s

∴ $S = ut + \frac{1}{2}at^{2} \\$ gives

$S = (0)(20) + \frac{1}{2}(1.1)(20)^{2}$

$S = \frac{1}{2}(1.1)(20)^{2}$

$S =$ 220m

This is the distance covered in the first 20s.

The train then proceeds at constant speed for 1100m.

Now, we will calculate the speed attained here

From

$v = u +at$

Where $v$ is the final velocity

Hence,

$v = 0 + 1.1(20)$

$v = 1.1(20)$

$v =$ 22 m/s

This is the constant speed attained when it proceeds for 1100m

The train then slows down at a rate of 2.2 m/s² until it stops

We can calculate the distance covered while slowing down from

$v^{2} = u^{2} + 2as$

The initial velocity, $u$ here will be the final velocity before it started slowing down

∴ $u$ = 22 m/s

The final velocity will be 0, since it came to a stop.

∴ $v$ = 0 m/s

$a = -$ 2.2 m/s² ( - indicates deceleration)

Hence,

$v^{2} = u^{2} + 2as$ gives

$0^{2} =22^{2} +2(-2.2)s$

$0=22^{2} - (4.4)s\\4.4s = 484\\s = \frac{484}{4.4} \\s = 110m$

This is the distance traveled while slowing down.

A) The distance between the stations is

220m + 1100m + 110m

= 1430m

Hence, the distance between the stations is 1430m

B) The time it takes the train to go between the stations

The time spent while accelerating at 1.1 m/s² is 20s

We will calculate the time spent when it proceeds at a constant speed of 22 m/s for 1100m,

From,

$Speed =\frac{Distance}{Time}\\$

Then,

$Time = \frac{Distance}{Speed}$

$Time = \frac{1100}{22}$

Time = 50 s

And then, the time spent while decelerating (that is, while slowing down)

From,

$v = u + at\\0 = 22 +(-2.2)t\\2.2t = 22\\t = \frac{22}{2.2} \\t= 10 s$

This is the time spent while slowing down until it stops at the station.

Hence, The time it takes the train to go between the stations is

20s + 50s + 10s = 80s

The time it takes the train to go between the stations is 80s

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3 years ago