Question #230b8

Let's begin by finding the acceleration for the first time period:

`a = (v_"final"-v_"initial")/(t_"final"-t_"initial")`

Substitute in the values:

`a = (44"ms"^-1-0"ms"^-1)/(4"s" - 0"s")`

Do the computation:

`a = 11"ms"^-2`

The distance for that interval is:

`d = 1/2a(t_"final"-t_"initial")^2+v_"initial"(t_"final"-t_"initial")`

Substitute in the values:

`d = 1/2(11"ms"^-2)(4"s"-0"s")^2+(0"ms"^-1)(4"s"-0"s")`

Do the computation:

`d = 88"m"`

Repeat this for the second interval:

`a = (v_"final"-v_"initial")/(t_"final"-t_"initial")`

Substitute in the values:

`a = (280"ms"^-1 -44"ms"^-1)/(12"s" - 4"s")`

Do the computation:

`a = 29.5"ms"^-2`

The distance for that interval is:

`d = 1/2a(t_"final"-t_"initial")^2+v_"initial"(t_"final"-t_"initial")`

Substitute in the values:

`d = 1/2(29.5"ms"^-2)(12"s"-4"s")^2+(44"ms"^-1)(12"s"-4"s")`

Do the computation:

`d = 1296"m"`

Add the two distances together: `d = 88"m" + 1296"m" = 1384"m"`

We need to consider both time intervals separately.

1. Time interval `0-4.0s`
   Using the following kinematic expression to calculate acceleration `a` during the interval

   `v=u+at` ..........(1)
   where `v` is the final velocity after time `t` and `u` is the initial velocity.

`44=0+a_1xx4`
`=>a_1=44/4=11ms^-2`

To calculate distance traveled `s_1` during this interval we use the other kinematic expression

`v^2-u^2=2as` ........(2)

`44^2-0^2=2xx11xxs_1`
`=>s_1=44^2/22=88m`
2. Time interval `4.0-12.0s`

Using equation (1) we get
`280=44+a_2xx8`
`=>a_1=(280-44)/8=29.5ms^-2`

Using equation (2) to calculate distance traveled `s_2` during this interval we get

`280^2-44^2=2xx29.5xxs_2`
`=>s_2=(280^2-44^2)/59=1296m`

Total distance traveled`=s_1+s_2=1384m`

It is given in the problem that

at `t=0s " velocity"(v_0)=0`

at `t=4s " velocity"(v_4)=44m"/"s`

at `t=12s " velocity"(v_12)=280m"/"s`

If we assume that the car travels 0 to 4s with a uniform acceleration and 4s to 12s with another uniform acceleration
,then the distance traveled during first 4 sec will be `S_1=(v_0+v_4)/2xx4=(0+44)/2xx4=88m`

and the distance traveled during last 8 sec will be `S_2=(v_4+v_12)/2xx8=(44+280)/2xx8=1296m`

Hence total distance traveled by the car will be

`S=S_1+S_2=88+1296=1384m`

`"area colored red :"A_r=(44*4)/2=88" "m`
`"area colored green:"A_g=((280+44)/2)*8=324*4=1296`
`"total area:"A_t=1296+88=1384" "m`