How I can find time by this formula S=ut+1/2at^2

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Homework Help Overview

The discussion revolves around solving a quadratic equation derived from the kinematic equation S = ut + 1/2at², where the original poster provides specific values for displacement, initial velocity, and acceleration. The goal is to find the time variable, t.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss transforming the equation into standard quadratic form and applying the quadratic formula. There are mentions of checking the discriminant (b² - 4ac) to determine the nature of the roots. Some participants express uncertainty about the implications of negative acceleration on the results.

Discussion Status

The conversation includes various interpretations of the quadratic equation and its solutions. Some participants offer guidance on using the quadratic formula, while others question the validity of certain solutions based on the physical context of the problem. There is no explicit consensus on the correct approach or answer.

Contextual Notes

Participants are navigating the implications of negative acceleration and its effect on the calculated time, as well as ensuring that the physical interpretation of time aligns with the mathematical results.

manal950
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Hi all

I have

s = 48 m
u = 30 m/s
a = -9.37

...

S = ut +1/2at^2

48 = 30t + 1/2( -9.37 ) t^2

Now after that , what should I do ?
 
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Algebra. This is a quadratic equation. Put it into standard form and then find the roots for t (i.e. used the quadratic formula, but with t instead the of the traditional x).
 
we havew done this sort of calc. Do you know the x =[-b +/- sqrt(b^2-4ac)]/2a equation ?
before you use this always check the b^2-4ac value. It tells you what sort of answer youwill get.
In this question I got b^2 - 4ac to be =1 which means there is only one answer and quite easy to work out (b/2a)
 
turn 48 = 30t + 1/2( -9.37 ) t^2 into 4.685t^2+30t-48=0 (turn a in positive) which is analog to ax^2+bx+c=0
using x =[-b +/- sqrt(b^2-4ac)]/2a you'll find: t1=1.33 and t2=-7.73.
the right answer is t1 bacause time can't be negative
 
thats wrong. If you use 1.33 and notice the - sign for acceleration...it does not give an equal answer.
30/9.37 = 3.2 is correct = (b/2a)
 

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