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Suppose , a particle is moving away with constant acceleration . It passes distance s by time t . A chart of their values is given in the following :

$$\begin{array}{|l|l|l|l|l|l|l|} \hline t & 0 & 1 & 3 & 6 & 8 & 10\\ \hline s & 0 & 10 & 42 & 120 & 192 & 280\\ \hline \end{array}$$

I have drawn a graph of t in x-axis and s in y-axis .The graph is as follows : enter image description here

I want to determine the initial velocity and acceleration from this graph . I know that initial velocity is the velocity at time t = 0 . And we can get velocity from the slope of s-t graph . But I can't determine initial velocity and acceleration .

Asaf Karagila
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Way to infinity
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1 Answers1

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Hint: Find $v$ and $a$ using your table if $s = vt+a\frac{t^2}{2}$.

njguliyev
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  • I cant use the formula rather I have to determine intial velocity and acceleration from graph . Can you plz see the question carefully ? – Way to infinity Aug 21 '13 at 10:35
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    I don't understand what do you mean by "from graph", but I can explain how to find $v$ and $a$ more "high-school-like", i.e. without using the abovementioned formula. Since your particle is moving with constant acceleration the velocity increases the same amount (which equals the acceleration) during every second. So the mean velocity between times $6$ and $8$ will be equal to the velocity at $t=7$: $v(7)=36$. By the same reasoning we have $v(9)=44$. Therefore $a=\frac{44-36}{2}=4$. Now you know the acceleration and the velocity at time $7$, so the initial velocity $v(0)=36-4 \cdot 7 = 8$. – njguliyev Aug 21 '13 at 13:48
  • Many many thanks for your valuable lecture .This comment is very useful for me to get solution of this problem . – Way to infinity Aug 22 '13 at 21:13
  • Well . I want to know another question . How can you tell that at t= 7 , the value of v(7) is 36 ? – Way to infinity Aug 22 '13 at 21:15
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    $v(7)$ is equal to the mean velocity between times $6$ and $8$: $$\frac{192-120}{2}.$$ – njguliyev Aug 22 '13 at 21:20