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.
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.
2 Answers
 Assuming that acceleration changes from
#t=4# to#t=6# linearly, we see that#dot a=(5.54.2)/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)  Assuming that acceleration changes from
#t=6# to#t=8# linearly, we see that for this time interval#dot a=(5.15.5)/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)  Assuming that acceleration changes from
#t=8# to#t=10# linearly, we see that#dot a=(6.25.1)/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)
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
Area under this graph is change in velocity of object

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"# 
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"# 
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