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#