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.
enter image source here

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

    dot a=(5.5-4.2)/2=0.65\ ms^-3
    Integrating both sides with respect to time we get
    a=0.65t+C ......(1)
    where C is a constant of integration to be evaluated from initial conditions. It is given that at t=4=>t_0=0, a=4.2. Inserting in (1) we get
    4.2=0.65xx0+C
    =>C=4.2
    Inserting in (1) we get
    a=0.65t+4.2 .......(2)
    We know that a=dot v. Integrating with respect to time both sides we get
    v=int(0.65t+1.6)\ dt
    =>v=0.65t^2/2+4.2t+C_1 .....(3)
    where C_1 is a constant of integration to be evaluated from initial conditions. Let us assume that at t=4=>t_0=0, v=v_4. Inserting in (3) we get
    =>v_4=0.65xx0^2/2+4.2xx0+C_1
    =>C_1=v_4
    Inserting in (3) we get
    =>v=0.325t^2+4.2t+v_4
    calculating velocity at t=6=> this acceleration was in operation for 2s
    v_6=0.325xx2^2+4.2xx2+v_4
    =>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=(5.1-5.5)/2=-0.2\ ms^-3
    Integrating both sides with respect to time we get
    a=-0.2t+C_2 ......(5)
    where C_2 is a constant of integration to be evaluated from initial conditions. It is given that at t=6=>t_0=0, a=5.5. Inserting in (5) we get
    5.5=-0.2xx0+C_2
    =>C_2=5.5
    Inserting in (5) we get
    a=-0.2t+5.5 .......(6)
    We know that a=dot v. Integrating with respect to time both sides we get
    v=int(-0.2t+5.5)\ dt
    =>v=-0.1t^2+5.5t+C_3 .....(7)
    where C_3 is a constant of integration to be evaluated from initial conditions. We have at t=6=>t_0=0, v_6=9.6+v_4. Inserting in (3) we get
    =>9.6+v_4=-0.1xx0^2+5.5xx0+C_3
    =>C_3=9.6+v_4
    Inserting in (7) we get
    =>v=-0.1t^2+5.5t+9.6+v_4
    calculating velocity at t=8=> this acceleration was in effect for 2s
    v_8=-0.1xx2^2+5.5xx2+9.6+v_4
    =>v_8=20.2+v_4 .........(8)

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

    dot a=(6.2-5.1)/2=0.55\ ms^-3
    Integrating both sides with respect to time we get
    a=0.55t+C_4 ......(9)
    where C_4 is a constant of integration to be evaluated from initial conditions. It is given that at t=8=>t_0=0, a=5.1. Inserting in (9) we get
    5.1=0.55xx0+C_4
    =>C_4=5.1
    Inserting in (9) we get
    a=0.55t+5.1 .......(10)
    We know that a=dot v. Integrating with respect to time both sides we get
    v=int(0.55t+5.1)\ dt
    =>v=0.55t^2/2+5.1t+C_5 .....(11)
    where C_5 is a constant of integration to be evaluated from initial conditions. We have calculated that at t=8=>t_0=0, v_8=20.2+v_4. Inserting in (11) we get
    =>20.2+v_4=0.55xx0^2/2+5.1xx0+C_5
    =>C_5=20.2+v_4
    Inserting in (11) we get
    =>v=0.275t^2+5.1t+20.2+v_4
    calculating velocity at t=10=> this acceleration was in operation for 2s
    v_10=0.275xx2^2+5.1xx2+20.4+v_4
    =>v_10=(31.7+v_4)\ ms^-1 .........(12)

May 29, 2018

"10.6 m/s"

Explanation:

Acceleration is given by

"a" = "Δv"/"t"

"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

enter image source here

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

"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"