Time Taken for Moving 200m with Constant Speed of 10m/s

[adjacent]

NOT a home work
A person moves with constant speed of 10m/s(Initially as well as finally). His acceleration is therefore zero.Distance moved is 200meter.Find time taken.I tried to use the equation below but had difficulty making t the subject
S=ut+1/2at²
a=Acceleration
t=time
u=speed
S=Distance moved
If I use the values given I could easily make t the subject
That is 200=10t+1/2*0*t²
Giving 200=10t (t=200/10) that is t=s/u in this case

I can't make t the subject using letters ONLY.There should be a way

 

[phinds]

If a man travels for 2 hours at 60 mph, how far does he go? What equation do you need to solve that question?

 

[tiny-tim]

hi adjacent!

S=ut+1/2at²
…
I can't make t the subject using letters ONLY.There should be a way

if a ≠ 0, this is a quadratic equation, which you can solve with t = -b ± √(b² - ) etc

if a = 0, it's simply s = ut, so t = s/u

 

[adjacent]

if a ≠ 0, this is a quadratic equation, which you can solve with t = -b ± √(b² - ) etc
if a = 0, it's simply s = ut, so t = s/u

S=ut+1/2at²
That means t cannot be made the subject of the formula by methods I use with (S=1/2(u+v)t)Etc?
What if S=ut+1/2at² is given and asked to solve for t?

 

[tiny-tim]

yes, solve it as a quadratic equation!

 

[adjacent]

S=ut+1/2at²

What if S=ut+1/2at² is given and asked to solve for t?

For example no values for a,u,tetc were given.I solve it as a quadratic,If a turn out to be 0 then It would be wrong.Right?Meaning the equation is wrong for non accelerating objects(when for t)but when solved for S it is right even for non accelerating objects.How is it?I need proof

 

[tiny-tim]

i don't understand

show us your solution for the quadratic​

 

[adjacent]

i don't understand

show us your solution for the quadratic​

For example,
a=0
s=200m
u=10m/s
t=20s
I can solve the equation even when a=0,(for s)
S=ut+1/2at²
=10*20+1/2*0*20² =200m
But for t,
0.5at²+ut-s=0
t=(-b+√(b²-4ac)/2a
t=(-10+√(10²-4(0)(-200))/(2*0) = MATH ERROR
In general If acceleration=0 - cannot solve for t
If acceleration≠0 - can solve for t
If acceleration is either ≠or= -can solve for S
Why?

 

[tiny-tim]

In general If acceleration=0 - cannot solve for t
If acceleration≠0 - can solve for t
If acceleration is either ≠or= -can solve for S
Why?

because the quadratic formula (-b ± √etc) does not work for a = 0

 

[adjacent]

because the quadratic formula (-b ± √etc) does not work for a = 0

But if it can be solved for S,There should be some method to solve for t too.(Except t=s/u which is obtained once you know the value of a)
Just Algebra(Only letters)?

 

[Mark44]

But if it can be solved for S,There should be some method to solve for t too.(Except t=s/u which is obtained once you know the value of a)
Just Algebra(Only letters)?

tiny-tim answered your question in post #3. There are two cases: a = 0 and a ≠ 0.

If a = 0, then t = s/v.
If a ≠ 0, then you can solve for t by using the quadratic formula.

 

[pwsnafu]

But if it can be solved for S,There should be some method to solve for t too.(Except t=s/u which is obtained once you know the value of a)
Just Algebra(Only letters)?

A quadratic equation means that a≠0.
$0x^2 + 2x + 4$ is not a quadratic.

 

[Mentallic]

But if it can be solved for S,There should be some method to solve for t too.(Except t=s/u which is obtained once you know the value of a)
Just Algebra(Only letters)?

I see what you're saying, and the answer to it is that the quadratic expression given by t = ... isn't quite complete because it assumes $a\neq 0$. What we need is a piecewise function to describe it

If $s = ut + 1/2at^2$

Then

$$t = <br /> \begin{cases}<br /> \frac{-u\pm \sqrt{u^2+2as}}{a}, & a\neq 0 \\<br /> \frac{s}{u}, & a=0<br /> \end{cases}$$

You need to also keep in mind that if the discriminant is less than zero, which is $a\neq 0$ but $u^2+2as<0$ then you won't have a real value for t. Physically, this means that the object won't ever cross the line at displacement s because it would be accelerating away from that direction.