What is the acceleration of this particle?

Question

A body is travelling along a straight line with constant acceleration $a\,{\mathrm{m\,s^{-2}}}$ . The body passes a fixed point $O$ on the line with velocity $u\,{\mathrm{m\,s^{-1}}}$ . Between $4\,\text{s}$ and $5\,\text{s}$ after passing $O$ the body travels $10\,\text{m}$ . Between $6\,\text{s}$ and $7\,\text{s}$ after passing $O$ the body travels $12\,\text{m}$ .

Calculate the values of $u$ and $a$ .

Solution

We are given distances at various times so the most useful equation of motion is going to be $s=ut+\frac{1}{2}at^2.$

We can use the equation to find the displacements at $5\,\text{s}$ and $4\,\text{s}$ . $s_5=5u+\frac{25}{2}a$ $s_4=4u+\frac{16}{2}a$ Subtracting gives us the distance travelled in this period which we are told is $10$ . We can do the same for the period $6\,\text{s}$ to $7\,\text{s}$ . $10=u+\frac{25-16}{2}a=u+4.5a$ $12=u+\frac{49-36}{2}a=u+6.5a$ We now can subtract these equations and solve for $a$ , $-2=-2a$ $a=1$ and substitute this value into one of the above equations to find $u$ . $10=u+4.5$ $u=5.5$

Alternatively, we could use a graphical approach to visually achieve the same solution. If we draw a velocity-time graph, the area underneath is the distance travelled.

From the information we are given, the difference between the two shaded areas is $2\,\text{m}$ . In the period of time between $4\,\text{s}$ and $6\,\text{s}$ the velocity increases by $2a$ . So the difference between the shaded areas is a rectangle of base $1$ and height $2a$ . Since we know this difference is $2$ we have $2a\times1=2$ $a=1$

Can you also deduce the value of $u$ from this diagram?

What is the acceleration of this particle?

Question

A body is travelling along a straight line with constant acceleration

a\,{\mathrm{m\,s^{-2}}}

. The body passes a fixed point

O

on the line with velocity

u\,{\mathrm{m\,s^{-1}}}

. Between

4\,\text{s}

and

5\,\text{s}

after passing

O

the body travels

10\,\text{m}

. Between

6\,\text{s}

and

7\,\text{s}

after passing

O

the body travels

12\,\text{m}

Calculate the values of

u

and

a

Solution

We are given distances at various times so the most useful equation of motion is going to be

s=ut+\frac{1}{2}at^2.

We can use the equation to find the displacements at

5\,\text{s}

and

4\,\text{s}

s_5=5u+\frac{25}{2}a

s_4=4u+\frac{16}{2}a

Subtracting gives us the distance travelled in this period which we are told is

10

. We can do the same for the period

6\,\text{s}

7\,\text{s}

10=u+\frac{25-16}{2}a=u+4.5a

12=u+\frac{49-36}{2}a=u+6.5a

We now can subtract these equations and solve for

a

-2=-2a

a=1

and substitute this value into one of the above equations to find

u

10=u+4.5

u=5.5

Alternatively, we could use a graphical approach to visually achieve the same solution. If we draw a velocity-time graph, the area underneath is the distance travelled.

From the information we are given, the difference between the two shaded areas is

2\,\text{m}

. In the period of time between

4\,\text{s}

and

6\,\text{s}

the velocity increases by

2a

. So the difference between the shaded areas is a rectangle of base

1

and height

2a

. Since we know this difference is

2

we have

2a\times1=2

a=1

Can you also deduce the value of

u

from this diagram?