Question

It is question about finding distance from velocity time graph?

Answer 1

See below.

Explanation:

If you have a graph of velocity against time, the area under the graph represents the distance.

This makes sense if you consider this:

`"Distance"="velocity"xx"time"`

On the graph:

Distance along t axis x distance along v axis.

You can find the area if it linear by using areas of squares, rectangles, trapeziums etc.

Answer 2

This is what I get

Explanation:

Let us start with Part
(ii) We need to use kinematic expression

`v=u+at`

Inserting given values

`V=-15+7.5xx(10-6)`
`=>V=-15+30=15\ ms^-1`

(iii)

1. Area of trapezoid `=1/2("sum of parallel bases")xx"height"`
   Length of top base `=3.5-1.5=2\ s`
   To calculate length of lower base we need to find time `t_1` when velocity becomes `=0`. Counting time from `t=3.5\ s`
   From the information given in (i)

   `0=10+(-10)xxt_1`
   `=>t_1=1\ s`
   `:.` Length of lower base`=3.5+1=4.5\ s`
   Now Area of trapezoid `=1/2(2+4.5)xx10=32.5\ m`

2. Area of first triangle `=1/2"base"xx"height"`
   To calculate length of base we need to find time `t_2` when velocity becomes `=0`. Counting time from `t=6\ s`
   From the information given in (ii)

   `0=-15+7.5xxt_2`
   `=>t_2=2\ s`
   `:.` Length of base `=` Time gap between two instances when velocity becomes zero`=8-4.5=3.5\ s`
   Area of first triangle `=1/2xx3.5xx|-15|=26.25\ m`
   We have taken the magnitude of velocity as we need to find distance which is a scalar quantity.
   (if we take negative sign in consideration we will get displacement which is a vector quantity)

3. From the given and calculated values
   Total distance `=100=`Area of trapezoid`+`Area of first triangle`+`Area of second triangle
   `=>100=32.5+26.25+`Area of second triangle
   `=>`Area of second triangle`=41.25\ m`
   `41.25=1/2"base"xx"height"`
   `=>41.25=1/2"base"xx15`
   `=>"base"=41.25xx2/15=5.5\ s`
   `:.T=8+5.5=13.5\ s`

Answer 3

`T=13.5\ s`

Explanation:

For question (ii), you have made an error.

Using `v=u+at`

Initial velocity `= -15ms^(-1)` `color(red)("at" \ \ \t=6)`

Acceleration `= 7.5ms^(-2)`

`t=10-6=4`

`v=-15+7.5(4)=15ms^(-1)`

Question (iii).

We need to find areas A , B, C

In order to do this we first need to find `t_1 and t_2`

For `t_1`

From question (i) we are told acceleration between 3.5 and 6 is `-10ms^(-1)`

Using `\ \ \ \v=u+at`

Velocity is zero at `t_1`

Velocity is `10ms^(-1)` at `t=3.5`

`:.`

`0=10-10(t_1-3.5)`

`0=10-10t_1+35`

`-45=-10t_1=>t_1=color(blue)(4.5)`

For `t_2`

We know from (ii), that the acceleration between `t=6 and t=10` is `7.5ms^(-2)`

Velocity at `t=10` is `15ms^-1` ( we found this before for (ii))

Velocity at `t_2=0`

Using `v=u+at`

`15=0+7.5(10-t_2)`

`15/7.5=10-t_2`

`t_2=color(blue)(8)`

Area of trapezium (trapezoid) A

`A=1/2(t_1+(3.5-1.5))(10)`

`A=1/2(4.5+2)(10)=32.5m^2`

Area of triangle B

`1/2(t_2-t_1)(15)`

`1/2(3.5)(15)=26.25m^2`

Area of triangle C

`1/2(T-t_2)(10)`

`1/2(T-8)(15)`

`7.5T-60m^2`

The sum of these areas is `100`

`:.`

`32.5+26.25+7.5T-60=100`

Solving for T:

`T=13.5 \ s`