# Motion Problem .

1. Jul 12, 2015

### Shahab Mirza

1. The problem statement, all variables and given/known data

A car travelling 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 .

3. 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

2. Jul 12, 2015

### Staff: Mentor

3. Jul 12, 2015

### 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!

4. Jul 12, 2015

### 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 isnt ? please make me clear or I am going fine?

5. Jul 12, 2015

### 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...

6. Jul 12, 2015

### Shahab Mirza

Yes Thanks , this was really a perfect solution , I was wondering how s = vt gonna solve this , Thanks alot

7. Jul 12, 2015

### Staff: Mentor

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

8. Jul 12, 2015

### Shahab Mirza

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

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

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9. Jul 12, 2015

### Physics Newbie

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

10. Jul 12, 2015

### Shahab Mirza

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 alot , Thanks

11. Jul 12, 2015

### Staff: Mentor

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

Your original approach using average velocity was just fine.

12. Jul 12, 2015

### Shahab Mirza

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