What is the change in the object’s velocity from t=4 to t=10?

The acceleration of an object at various times is shown in the table below. Using the information in the table, estimate the change in the object’s velocity from t=4 to t=10.

May 26, 2018
1. Assuming that acceleration changes from $t = 4$ to $t = 6$ linearly, we see that

$\dot{a} = \frac{5.5 - 4.2}{2} = 0.65 \setminus m {s}^{-} 3$
Integrating both sides with respect to time we get
$a = 0.65 t + C$ ......(1)
where $C$ is a constant of integration to be evaluated from initial conditions. It is given that at $t = 4 \implies {t}_{0} = 0 , a = 4.2$. Inserting in (1) we get
$4.2 = 0.65 \times 0 + C$
$\implies C = 4.2$
Inserting in (1) we get
$a = 0.65 t + 4.2$ .......(2)
We know that $a = \dot{v}$. Integrating with respect to time both sides we get
$v = \int \left(0.65 t + 1.6\right) \setminus \mathrm{dt}$
$\implies v = 0.65 {t}^{2} / 2 + 4.2 t + {C}_{1}$ .....(3)
where ${C}_{1}$ is a constant of integration to be evaluated from initial conditions. Let us assume that at $t = 4 \implies {t}_{0} = 0 , v = {v}_{4}$. Inserting in (3) we get
$\implies {v}_{4} = 0.65 \times {0}^{2} / 2 + 4.2 \times 0 + {C}_{1}$
$\implies {C}_{1} = {v}_{4}$
Inserting in (3) we get
$\implies v = 0.325 {t}^{2} + 4.2 t + {v}_{4}$
calculating velocity at $t = 6 \implies$ this acceleration was in operation for $2 s$
${v}_{6} = 0.325 \times {2}^{2} + 4.2 \times 2 + {v}_{4}$
$\implies {v}_{6} = 9.7 + {v}_{4}$ .........(4)

2. Assuming that acceleration changes from $t = 6$ to $t = 8$ linearly, we see that for this time interval

$\dot{a} = \frac{5.1 - 5.5}{2} = - 0.2 \setminus m {s}^{-} 3$
Integrating both sides with respect to time we get
$a = - 0.2 t + {C}_{2}$ ......(5)
where ${C}_{2}$ is a constant of integration to be evaluated from initial conditions. It is given that at $t = 6 \implies {t}_{0} = 0 , a = 5.5$. Inserting in (5) we get
$5.5 = - 0.2 \times 0 + {C}_{2}$
$\implies {C}_{2} = 5.5$
Inserting in (5) we get
$a = - 0.2 t + 5.5$ .......(6)
We know that $a = \dot{v}$. Integrating with respect to time both sides we get
$v = \int \left(- 0.2 t + 5.5\right) \setminus \mathrm{dt}$
$\implies v = - 0.1 {t}^{2} + 5.5 t + {C}_{3}$ .....(7)
where ${C}_{3}$ is a constant of integration to be evaluated from initial conditions. We have at $t = 6 \implies {t}_{0} = 0 , {v}_{6} = 9.6 + {v}_{4}$. Inserting in (3) we get
$\implies 9.6 + {v}_{4} = - 0.1 \times {0}^{2} + 5.5 \times 0 + {C}_{3}$
$\implies {C}_{3} = 9.6 + {v}_{4}$
Inserting in (7) we get
$\implies v = - 0.1 {t}^{2} + 5.5 t + 9.6 + {v}_{4}$
calculating velocity at $t = 8 \implies$ this acceleration was in effect for $2 s$
${v}_{8} = - 0.1 \times {2}^{2} + 5.5 \times 2 + 9.6 + {v}_{4}$
$\implies {v}_{8} = 20.2 + {v}_{4}$ .........(8)

3. Assuming that acceleration changes from $t = 8$ to $t = 10$ linearly, we see that

$\dot{a} = \frac{6.2 - 5.1}{2} = 0.55 \setminus m {s}^{-} 3$
Integrating both sides with respect to time we get
$a = 0.55 t + {C}_{4}$ ......(9)
where ${C}_{4}$ is a constant of integration to be evaluated from initial conditions. It is given that at $t = 8 \implies {t}_{0} = 0 , a = 5.1$. Inserting in (9) we get
$5.1 = 0.55 \times 0 + {C}_{4}$
$\implies {C}_{4} = 5.1$
Inserting in (9) we get
$a = 0.55 t + 5.1$ .......(10)
We know that $a = \dot{v}$. Integrating with respect to time both sides we get
$v = \int \left(0.55 t + 5.1\right) \setminus \mathrm{dt}$
$\implies v = 0.55 {t}^{2} / 2 + 5.1 t + {C}_{5}$ .....(11)
where ${C}_{5}$ is a constant of integration to be evaluated from initial conditions. We have calculated that at $t = 8 \implies {t}_{0} = 0 , {v}_{8} = 20.2 + {v}_{4}$. Inserting in (11) we get
$\implies 20.2 + {v}_{4} = 0.55 \times {0}^{2} / 2 + 5.1 \times 0 + {C}_{5}$
$\implies {C}_{5} = 20.2 + {v}_{4}$
Inserting in (11) we get
$\implies v = 0.275 {t}^{2} + 5.1 t + 20.2 + {v}_{4}$
calculating velocity at $t = 10 \implies$ this acceleration was in operation for $2 s$
${v}_{10} = 0.275 \times {2}^{2} + 5.1 \times 2 + 20.4 + {v}_{4}$
$\implies {v}_{10} = \left(31.7 + {v}_{4}\right) \setminus m {s}^{-} 1$ .........(12)

May 29, 2018

$\text{10.6 m/s}$

Explanation:

Acceleration is given by

$\text{a" = "Δv"/"t}$

$\text{Δv = at}$

From above equation we can say that area of an acceleration vs time graph gives us change in velocity.

Acceleration vs Time graph from $4 - 10$ seconds is shown below

Area under this graph is change in velocity of object

1. From $t = 4$ to $t = 6$
"Δv"_1 = [1/2 × 2 × (5.5 - 4.2)] + [(6 - 4) × (4.2 - 3.5)]
color(white)(Δv_1) = 1.3 + 1.4
color(white)(Δv_1) = "2.7 m/s"

2. From $t = 6$ to $t = 8$
"Δv"_2 = [1/2 × 2 × (5.5 - 5.1)] + [(8 - 6) × (5.1 - 3.5)]
color(white)(Δv_2) = 0.4 + 3.2
color(white)(Δv_2) = "3.6 m/s"

3. From $t = 8$ to $t = 10$
"Δv"_3 = [1/2 × 2 × (6.2 - 5.1)] + [(10 - 8) × (5.1 - 3.5)]
color(white)(Δv_3) = 1.1 + 3.2
color(white)(Δv_3) = 4.3\ "m/s"

Change in velocity from $t = 4$ to $t = 10$ is

${\text{Δv = Δv"_1 + "Δv"_2 + "Δv}}_{3}$

color(white)(Δv) = "2.7 m/s + 3.6 m/s + 4.3 m/s"

color(white)(Δv) = "10.6 m/s"