#2**+8 **

Let the distance travelled in the fist 5 seconds be x metres

We know u=0

When t=5, s=x

When t=10 s=x+150

I am going to use the formula $${\mathtt{s}} = {\mathtt{ut}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{{\mathtt{at}}}^{{\mathtt{2}}}$$

since u=0, the formula becomes $${\mathtt{s}} = {\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{{\mathtt{at}}}^{{\mathtt{2}}}$$

Now we have 2 equations that can be solved simultaneously.

$$\\x=\frac{1}{2}a\times 5^2 \\\\\

x= \frac{1}{2}a\times 25 \qquad (1)\\\\

and\\\\

x+150=\frac{1}{2}a\times 10^2 \\\\

x+150=\frac{1}{2}a\times 100 \\\\

x+150=\frac{1}{2}a\times 25 \times 4 \\\\

\frac{x+150}{4}=\frac{1}{2}a\times 25 \qquad (2)\\\\

$putting these together I get\\\\

\begin{array}{rll}

x&=&\frac{x+150}{4}\\

4x&=&x+150\\

3x&=&150\\

x&=&50\\

\end{array}$$

So the ball rolls 50m in the first 5 seconds

Melody
Sep 27, 2014

#2**+8 **

Best Answer

Let the distance travelled in the fist 5 seconds be x metres

We know u=0

When t=5, s=x

When t=10 s=x+150

I am going to use the formula $${\mathtt{s}} = {\mathtt{ut}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{{\mathtt{at}}}^{{\mathtt{2}}}$$

since u=0, the formula becomes $${\mathtt{s}} = {\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{{\mathtt{at}}}^{{\mathtt{2}}}$$

Now we have 2 equations that can be solved simultaneously.

$$\\x=\frac{1}{2}a\times 5^2 \\\\\

x= \frac{1}{2}a\times 25 \qquad (1)\\\\

and\\\\

x+150=\frac{1}{2}a\times 10^2 \\\\

x+150=\frac{1}{2}a\times 100 \\\\

x+150=\frac{1}{2}a\times 25 \times 4 \\\\

\frac{x+150}{4}=\frac{1}{2}a\times 25 \qquad (2)\\\\

$putting these together I get\\\\

\begin{array}{rll}

x&=&\frac{x+150}{4}\\

4x&=&x+150\\

3x&=&150\\

x&=&50\\

\end{array}$$

So the ball rolls 50m in the first 5 seconds

Melody
Sep 27, 2014