Find how fast is the train travelling at the end of the fourth minute?

[chwala]

Homework Statement

See attached.

Relevant Equations

$s=vt$

Find the questions and its solution below;

Now would i be correct to use;
$v= u +at$ where i considered $v=33.3$m/s and $u=16.67$m/s as the given average velocities between the two points in reference. Then it follows that,
$33.33=16.67 + 150a$
$→a= 0.11$
We are told that $a$ is constant throughout... Letting the unknown final velocity (which we are looking for) $=v_2$, then
using $v=u+at$
$v_2 = 33.33 +(0.11×30)=33.33+3.3=36.63$m/s

 

[PeroK]

Homework Statement:: See attached.
Relevant Equations:: $s=vt$

Find the questions and its solution below;

View attachment 294244
View attachment 294245

Now would i be correct to use;
$v= u +at$ where i considered $v=33.3$m/s and $u=16.67$m/s as the given average velocities between the two points in reference. Then it follows that,
$33.33=16.67 + 150a$
$→a= 0.11$
We are told that $a$ is constant throughout... Letting the unknown final velocity (which we are looking for) $=v_2$, then
using $v=u+at$
$v_2 = 33.33 +(0.11×30)=33.33+3.3=36.63$m/s

Yes, you can do it using average speed. You know the speed after 2 minutes is $1.5$ km/min. And the speed at $3.5$ minutes is $2$ km/min. That gives $a = \frac 1 3$ (km/min)/min.

There is no need to convert to SI units. As long as the units are consistent throughout.

 

[mjc123]

Your answer is slightly wrong; I suspect you have made a mistake in converting to SI units. What are the given speeds and times for which you get 33.33 m/s and 16.67 m/s as "the given average velocities between the two points in reference"?

 

[chwala]

Your answer is slightly wrong; I suspect you have made a mistake in converting to SI units. What are the given speeds and times for which you get 33.33 m/s and 16.67 m/s as "the given average velocities between the two points in reference"?

True, i made a mistake on $u$, i would amend as follows;

$v= u +at$ where i considered $v=33.3$m/s and $u=22.2$m/s as the given average velocities between the two points in reference. Then it follows that,
$33.33=22.2 + 150a$
$→a= 0.074$
We are told that $a$ is constant throughout... Letting the unknown final velocity (which we are looking for) $=v_2$, then
using $v=u+at$
$v_2 = 33.33 +(0.074×30)=33.33+2.2=35.55$m/s

 

[mjc123]

I asked what were the given speeds and times, i.e. the quoted values in km/min and min that you have converted?

 

[chwala]

I asked what were the given speeds and times, i.e. the quoted values in km/min and min that you have converted?

ok, i converted $\frac {4}{3}$ km/min into approximately $22.2222$ m/s and also converted the $\frac {2}{1}$ km/min to $33.3333$ m/s

 

[mjc123]

And the times?

 

[chwala]

The times between the two velocities is $150$ and since we are told that acceleration is constant, then i used that in deducing the final $30$ seconds of the journey.

 

[mjc123]

No I didn't; don't put your words in my mouth.

What is the difference between 1.5 min and 3.5 min in seconds?

 

[chwala]

No I didn't; don't put your words in my mouth.

What is the difference between 1.5 min and 3.5 min in seconds?

$t=120$...oops, therefore,
$v= u +at$ where i considered $v=33.3$m/s and $u=22.2$m/s as the given average velocities between the two points in reference. Then it follows that,
$33.33=22.2 + 120a$
$→a= 0.092594$
We are told that $a$ is constant throughout... Letting the unknown final velocity (which we are looking for) $=v_2$, then
using $v=u+at$
$v_2 = 33.33 +(0.092594×30)=33.33+2.777=36.1111$m/s≡$2$$\frac{1}{6}$km/min
Thanks man...will counter check my working in future and minimize the errors...