Motion Problem Constant Acceleration from 10 to 50 m/s in 2 secs

[Shahab Mirza]

Homework Statement

A car traveling with constant acceleration increases its speed from 10 m/s to 50 m/s over a distance of 60 m . how long does this take?

a. 2 seconds b. 4 seconds .

The Attempt at a Solution

Data :
Avg Velocity = 10 + 50 / 2 = 30m/s
Distance = 60 meter
time = ?

Solution :

S= V x T

making "T" subject . Then

T= S/V
T= 60/30

T = 2 seconds . This Answer is correct , but please tell me that the way I have solved this is correct or not? Thanks

 

[Doc Al]

Your solution is fine.

 

[TSny]

I agree with Doc Al. Your method is correct.

But I'm in a nit picky mood this morning. Use parenthesis in the expression for calculating the average velocity in order to show that the 10 and 50 are added together first before dividing by 2:

( 10 + 50 ) / 2.

Anyway, good work!

 

[Shahab Mirza]

Thanks , But I am not satisfied with my solution, u know why sir? because S = vt is possible when initial velocity is 0 means herz equation s = vit + 1/2 a t^2 , but this eq can only be reduced to s =vt when acceleration is unity and initial velocity is 0 , in my case it isn't ? please make me clear or I am going fine?

 

[Physics Newbie]

Check this:
We have three equations of motion for constant acceleration along 1D,
1. v=u+at,
2. v*v - u*u=2as, and
3. S=ut + 1/2at*t

okay

Here u(initial velocity)=10m/s, v(final velocity)=50m/s, a=constant (so apply equations as it is a 1D problem), s=60, t=?
Using eq. 2 we get
2500-100=2*a*60
a=20m/s*s

Now using this value of 'a' in eq 1 we get;

50=10+20*t
t=2s

Thats it fella...

 

[Shahab Mirza]

Check this:
We have three equations of motion for constant acceleration along 1D,
1. v=u+at,
2. v*v - u*u=2as, and
3. S=ut + 1/2at*t

okay

Here u(initial velocity)=10m/s, v(final velocity)=50m/s, a=constant (so apply equations as it is a 1D problem), s=60, t=?
Using eq. 2 we get
2500-100=2*a*60
a=20m/s*s

Now using this value of 'a' in eq 1 we get;

50=10+20*t
t=2s

Thats it fella...

Yes Thanks , this was really a perfect solution , I was wondering how s = vt going to solve this , Thanks a lot

 

[Doc Al]

Thanks , But I am not satisfied with my solution, u know why sir? because S = vt is possible when initial velocity is 0 means herz equation s = vit + 1/2 a t^2 , but this eq can only be reduced to s =vt when acceleration is unity and initial velocity is 0 , in my case it isn't ? please make me clear or I am going fine?

What you're using, whether you realize it or not, is $\Delta S = v_{ave}t$. Which is fine here.

 

[Shahab Mirza]

What you're using, whether you realize it or not, is $\Delta S = v_{ave}t$. Which is fine here.

Thanks , but the sir who replied above gave this solution.

Here u(initial velocity)=10m/s, v(final velocity)=50m/s, a=constant (so apply equations as it is a 1D problem), s=60, t=?
Using eq. 2 we get
2500-100=2*a*60
a=20m/s*s

Now using this value of 'a' in eq 1 we get;

50=10+20*t
t=2sI think it was simple nasic problem, I followed question wrong i think.

 

[Physics Newbie]

Top tip:Whenever there is constant acc in 1D go for the equations...

 

[Shahab Mirza]

Top tip:Whenever there is constant acc in 1D go for the equations...

Its very kind of you , Thanks for sharing knowledge , actually I am going to give medical college admission test this year , it has 3 subjects on which test will be based , 1 BIology . 2 Physics 3 Chemistry . so these little concepts will help me a lot , Thanks

 

[Doc Al]

Thanks , but the sir who replied above gave this solution.

Here u(initial velocity)=10m/s, v(final velocity)=50m/s, a=constant (so apply equations as it is a 1D problem), s=60, t=?
Using eq. 2 we get
2500-100=2*a*60
a=20m/s*s

Now using this value of 'a' in eq 1 we get;

50=10+20*t
t=2s

There's nothing wrong with this solution, of course. But note the extra steps involved.

I think it was simple nasic problem, I followed question wrong i think.

Your original approach using average velocity was just fine.

 

[Shahab Mirza]

There's nothing wrong with this solution, of course. But note the extra steps involved.Your original approach using average velocity was just fine.

Yes I understood Sir , Thanks , I will keep all these things in my knowledge . Thanks