Feb 27, 2018

$a = 1 \frac{m}{s} ^ 2 \mathmr{and} u = 5.5 \frac{m}{s}$

Explanation:

With constant acceleration during a time interval $\Delta t$, the body's instantaneous velocity equals its average velocity when time is halfway through the time interval.

The body travels 10 m during the 1 s interval between t = 4 s and t = 5 s. The average velocity during that interval is

${V}_{\text{ave}} = 10 \frac{m}{1} s = 10 \frac{m}{s}$

Therefore the body's instantaneous velocity at t = 4.5 s was 10 m/s.

Using that same logic, the body's instantaneous velocity at t = 6.5 s was 12 m/s. We can now calculate acceleration.

$a = \frac{\Delta V}{\Delta t} = \frac{12 \frac{m}{s} - 10 \frac{m}{s}}{6.5 s - 4.5 s} = 1 \frac{m}{s} ^ 2$

Now the kinematic formula $v = u + a \cdot t$ will allow us to calculate u. I will use the time t = 4.5, therefore v will be 10 m/s.

$v = u + a \cdot t$

$10 \frac{m}{s} = u + 1 \frac{m}{s} ^ 2 \cdot 4.5 s$

$u = 10 \frac{m}{s} - 1 \frac{m}{s} ^ 2 \cdot 4.5 s = 10 \frac{m}{s} - 4.5 \frac{m}{s} = 5.5 \frac{m}{s}$

I hope this helps,
Steve

Feb 27, 2018

$\text{The answers are:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \setminus a = 1 \setminus \setminus \frac{m}{s} ^ 2 \setminus q \quad \setminus q \quad \setminus \text{and} \setminus q \quad \setminus q \quad v = \frac{11}{2} \setminus \setminus \frac{m}{s} \setminus \quad .$

Explanation:

$\text{For a body moving with constant acceleration" \ a, "along a straight line, the equation of motion -- }$
$\text{the position function" \ s(t), "is given by:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad s \left(t\right) \setminus = \setminus \frac{1}{2} a {t}^{2} + {v}_{0} t + {s}_{0} , \setminus q \quad \text{where:} \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \left(I\right)$

$\setminus \quad a = \text{acceleration," quad v_0 = "initial velocity" quad s_0 = "initial position.}$

$\text{[I don't know if you are allowed to assume this (!!) It is a}$
$\text{standard result for this type of motion. ]}$

$\text{At any rate, back to the above equation.}$

$\text{We can take the origin of the straight line to be a location of}$
$\text{our choice on the line. Taking the point O as the origin, will}$
$\text{prove to be very nice. Let's also start the clock when the body}$
$\text{is at point O.}$

$\text{So we have:}$

 \quad "position at time" \ t = 0 \ \ "is:" \quad "point O"; \qquad rArr \quad \ s_0 = 0.

 \quad "velocity at time" \ t = 0 \ \ "is:" \quad "velocity at point O"; \qquad rArr \quad v_0 = u.

$\text{Putting these into our equation of motion in (I), we have:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \setminus s \left(t\right) \setminus = \setminus \frac{1}{2} a {t}^{2} + {v}_{0} t + {s}_{0.}$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus s \left(t\right) \setminus = \setminus \frac{1}{2} a {t}^{2} + u t + 0$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus s \left(t\right) \setminus = \setminus \frac{1}{2} a {t}^{2} + u t . \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \left(I I\right)$

$\text{Now we are given: between 4 s and 5 s, the body traveled 10m.}$

 \qquad \qquad :. \qquad \qquad \qquad \qquad \qquad s(5) - s(4) \ = \ 10. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (III)

$\text{Now we are also given: between 6 s and 7 s, body traveled 12m.}$

 \qquad \qquad :. \qquad \qquad \qquad \qquad \qquad s(7) - s(6) \ = \ 12. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ (IV)

$\text{Using the equation of motion from (II), eqn. (III) becomes:}$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus \left[\frac{1}{2} a {\left(5\right)}^{2} + u \left(5\right)\right] - \left[\frac{1}{2} a {\left(4\right)}^{2} + u \left(4\right)\right] \setminus = \setminus 10.$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus \quad \left[\frac{25}{2} a + 5 u\right] - \left[\frac{16}{2} a + 4 u\right] \setminus = \setminus 10.$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \frac{9}{2} a + u \setminus = \setminus 10.$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \setminus 9 a + 2 u \setminus = \setminus 20. \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \setminus \setminus \left(V\right)$

$\text{Similarly, using it with eqn. (IV), we have:}$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus \left[\frac{1}{2} a {\left(7\right)}^{2} + u \left(7\right)\right] - \left[\frac{1}{2} a {\left(6\right)}^{2} + u \left(6\right)\right] \setminus = \setminus 12.$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus \quad \left[\frac{49}{2} a + 7 u\right] - \left[\frac{36}{2} a + 6 u\right] \setminus = \setminus 12.$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \frac{13}{2} a + u \setminus = \setminus 12.$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus 13 a + 2 u \setminus = \setminus 24. \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \setminus \left(V I\right)$

$\text{So now we see eqns. (V) and (VI) form a" \ \ 2 xx 2 \ "system of}$
$\text{linear equations for the desired quantities:" \quad a, u.}$

$\text{So, let's solve these:}$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \setminus 9 a + 2 u \setminus = \setminus 20.$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \setminus 13 a + 2 u \setminus = \setminus 24.$

$\text{(VI)" \ - \ (V):} \setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus 4 a \setminus = \setminus 4.$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad a \setminus = \setminus 1. \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \setminus \left(V I I\right)$

$\text{Substituting" \ a = 1 \ "into eqn. (V), we get:}$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \setminus 9 \left(1\right) + 2 u \setminus \setminus = \setminus 20.$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad 9 + 2 u \setminus = \setminus 20.$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \quad \setminus \setminus \setminus 2 u \setminus = \setminus 11.$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \setminus \setminus q \quad \setminus q \quad \setminus \quad \setminus u \setminus = \setminus \frac{11}{2.} \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \left(V I I\right)$

$\text{And so we have our desired results:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad a = 1 \setminus \setminus \frac{m}{s} ^ 2 \setminus q \quad \setminus q \quad \setminus \text{and} \setminus q \quad \setminus q \quad v = \frac{11}{2} \setminus \setminus \frac{m}{s} \setminus \quad .$