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...