Question #230b8

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Question #230b8

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Answer 1 · 2017-05-21T15:26:11

The total distance traveled is $1384 m$ $1384"m"$

Explanation:

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

$a = \frac{v_{final} - v_{initial}}{t_{final} - t_{initial}}$ $a = (v_"final"-v_"initial")/(t_"final"-t_"initial")$

Substitute in the values:

$a = \frac{44 {ms}^{- 1} - 0 {ms}^{- 1}}{4 s - 0 s}$ $a = (44"ms"^-1-0"ms"^-1)/(4"s" - 0"s")$

Do the computation:

$a = 11 {ms}^{- 2}$ $a = 11"ms"^-2$

The distance for that interval is:

$d = \frac{1}{2} a {(t_{final} - t_{initial})}_{final}^{2} + v_{initial} (t_{final} - t_{initial})$ $d = 1/2a(t_"final"-t_"initial")^2+v_"initial"(t_"final"-t_"initial")$

Substitute in the values:

$d = \frac{1}{2} (11 {ms}^{- 2}) {(4 s - 0 s)}^{2} + (0 {ms}^{- 1}) (4 s - 0 s)$ $d = 1/2(11"ms"^-2)(4"s"-0"s")^2+(0"ms"^-1)(4"s"-0"s")$

Do the computation:

$d = 88 m$ $d = 88"m"$

Repeat this for the second interval:

$a = \frac{v_{final} - v_{initial}}{t_{final} - t_{initial}}$ $a = (v_"final"-v_"initial")/(t_"final"-t_"initial")$

Substitute in the values:

$a = \frac{280 {ms}^{- 1} - 44 {ms}^{- 1}}{12 s - 4 s}$ $a = (280"ms"^-1 -44"ms"^-1)/(12"s" - 4"s")$

Do the computation:

$a = 29.5 {ms}^{- 2}$ $a = 29.5"ms"^-2$

The distance for that interval is:

$d = \frac{1}{2} a {(t_{final} - t_{initial})}_{final}^{2} + v_{initial} (t_{final} - t_{initial})$ $d = 1/2a(t_"final"-t_"initial")^2+v_"initial"(t_"final"-t_"initial")$

Substitute in the values:

$d = \frac{1}{2} (29.5 {ms}^{- 2}) {(12 s - 4 s)}^{2} + (44 {ms}^{- 1}) (12 s - 4 s)$ $d = 1/2(29.5"ms"^-2)(12"s"-4"s")^2+(44"ms"^-1)(12"s"-4"s")$

Do the computation:

$d = 1296 m$ $d = 1296"m"$

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

Answer 2 · 2017-05-21T16:03:14

We need to consider both time intervals separately.

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

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

$44 = 0 + a_{1} \times 4$ $44=0+a_1xx4$
$\Rightarrow a_{1} = \frac{44}{4} = 11 m s^{- 2}$ $=>a_1=44/4=11ms^-2$

To calculate distance traveled $s_{1}$ $s_1$ during this interval we use the other kinematic expression

$v^{2} - u^{2} = 2 a s$ $v^2-u^2=2as$ ........(2)

$44^{2} - 0^{2} = 2 \times 11 \times s_{1}$ $44^2-0^2=2xx11xxs_1$
$\Rightarrow s_{1} = \frac{44^{2}}{22} = 88 m$ $=>s_1=44^2/22=88m$
2. Time interval $4.0 - 12.0 s$ $4.0-12.0s$

Using equation (1) we get
$280 = 44 + a_{2} \times 8$ $280=44+a_2xx8$
$\Rightarrow a_{1} = \frac{280 - 44}{8} = 29.5 m s^{- 2}$ $=>a_1=(280-44)/8=29.5ms^-2$

Using equation (2) to calculate distance traveled $s_{2}$ $s_2$ during this interval we get

$280^{2} - 44^{2} = 2 \times 29.5 \times s_{2}$ $280^2-44^2=2xx29.5xxs_2$
$\Rightarrow s_{2} = \frac{280^{2} - 44^{2}}{59} = 1296 m$ $=>s_2=(280^2-44^2)/59=1296m$

Total distance traveled $= s_{1} + s_{2} = 1384 m$ $=s_1+s_2=1384m$

Answer 3 · 2017-05-21T16:30:10

It is given in the problem that

at $t = 0 s velocity (v_{0}) = 0$ $t=0s " velocity"(v_0)=0$

at $t = 4 s velocity (v_{4}) = 44 m / s$ $t=4s " velocity"(v_4)=44m"/"s$

at $t = 12 s velocity (v_{12}) = 280 m / s$ $t=12s " velocity"(v_12)=280m"/"s$

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 $S_{1} = \frac{v_{0} + v_{4}}{2} \times 4 = \frac{0 + 44}{2} \times 4 = 88 m$ $S_1=(v_0+v_4)/2xx4=(0+44)/2xx4=88m$

and the distance traveled during last 8 sec will be $S_{2} = \frac{v_{4} + v_{12}}{2} \times 8 = \frac{44 + 280}{2} \times 8 = 1296 m$ $S_2=(v_4+v_12)/2xx8=(44+280)/2xx8=1296m$

Hence total distance traveled by the car will be

$S = S_{1} + S_{2} = 88 + 1296 = 1384 m$ $S=S_1+S_2=88+1296=1384m$

Answer 4 · 2017-05-22T08:19:58

$A graphical solution.$ $"A graphical solution."$

Explanation:

enter image source here

$area colored red : A_{r} = \frac{44 \cdot 4}{2} = 88 m$ $"area colored red :"A_r=(44*4)/2=88" "m$
$area colored green: A_{g} = (\frac{280 + 44}{2}) \cdot 8 = 324 \cdot 4 = 1296$ $"area colored green:"A_g=((280+44)/2)*8=324*4=1296$
$total area: A_{t} = 1296 + 88 = 1384 m$ $"total area:"A_t=1296+88=1384" "m$

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