Let us assume at t = 0 a cart is leaving the origin with zero initial
velocity and constant acceleration of 3 m/s^{2}
in the positive x-direction.

Then a = ∆v/∆t, v_{f}
- v_{i} = a(t_{f} - t_{i}), and since v_{i} = t_{i}
= 0 we have v(t) = a*t = (3 m/s^{2})*t.

The velocity (m)/s versus time (s) graph is a straight line with is a
straight line passing through the origin with a slope of 3.

The acceleration
(m/s^{2}) versus time (s) graph yields a
straight line (a = 3) with zero slope.

Let us consider one-dimensional motion in the x-direction.

Let us make
this explicit by using the subscript x.

The average acceleration equals the instantaneous acceleration.
From

a_{x} = (v_{xf} - v_{xi})/(t_{f} - t_{i})

we obtain

a_{x}(t_{f }- t_{i}) = (v_{xf }- v_{xi}),

or

v_{xf }= v_{xi }+ a_{x}∆t.

The velocity versus time graph is a straight line.

The average velocity in a
time interval ∆t therefore is just the sum of the final and the initial
velocities divided by 2,

v_{x(avg) }= (v_{xf }+ v_{xi})/2.

The displacement is ∆x = v_{x(avg)}∆t. We can rewrite this
expression to obtain x_{f }- x_{i }= ½ (v_{xf }+ v_{xi})∆t,
or

x_{f }- x_{i }= v_{xi}∆t
+ ½a_{x}∆t^{2}.

We can also express the velocity as a function of the displacement.

∆x = ½(v_{xf }+ v_{xi})∆t = ½ (v_{xf }+ v_{xi})(v_{xf
}- v_{xi})/a_{x }= (v_{xf}^{2 }- v_{xi}^{2})/(2a_{x})
yields

v_{xf}^{2 }= v_{xi}^{2
}+ 2a_{x}(x_{f }- x_{i}).

The equations in red are the kinematic equations for motion in the x-direction with constant acceleration.

Choose your coordinates so that x_{i} = t_{i} = 0. Then x = v_{i}t +
½at^{2}.

This is the equation of a parabola. The
position versus time graph is a section of a parabola.

In the limit a = 0
it becomes a straight line.

The speed versus time graph on the right represents the motion of a car. Approximately how far did the car travel during the first 5 seconds?

Solution:

The speed versus time graph is a straight line.

We have motion with
constant acceleration.

The slope of the graph represents the
acceleration.

a_{x} = (v_{xf} - v_{xi})/(t_{f}
- t_{i}) =
(-40 m/s)/(10 s ) = -4 m/s^{2}.

For motion with constant acceleration we have

∆x_{i
}= v_{xi}∆t
+ ½a_{x}∆t^{2}.

After 5 s we have

∆x_{
}=
40m/s * 5 s - ½(4 m/s^{2})*(5
s)^{2} = 150 m.

During the first 5 seconds the car traveled 150 m.

A particle is moving with velocity **v**_{0 }= 60** i **(m/s)
at t = 0. Between t = 0 and t = 15 s the velocity decreases uniformly to zero.
What was the acceleration during this 15 s time interval? What is the
significance of the sign in your answer?

Solution:

Since the velocity decreases uniformly, the acceleration is constant. We
therefore have a_{x} = (v_{xf} - v_{xi})/∆t = (0 - 60
m/s)/15 s = -4m/s^{2}.

The minus sign tells us that the acceleration vector is pointing in the negative
x-direction. The velocity and the acceleration vector point in opposite
directions. The particle is slowing down.

Blood is accelerated from rest to v = 30.0 cm/s in a distance of 1.80 cm by
the left ventricle of the heart. Assume constant acceleration.

(a) Find the acceleration a.

(b) For how long does the blood accelerate?

(c) Is the answer reasonable when compared with the time for a heartbeat?

Solution:

(a) Given: v_{i} = 0, v_{f} = 0.3 m/s,
∆v = 0.3 m/s.

x_{i} = 0, x_{f} = 0.018m, ∆x_{
}=
0.018 m.

Kinematic equation: v_{xf}^{2
}= v_{xi}^{2
}+ 2a_{x}(x_{f }- x_{i})

Solve for a_{x} = (v_{xf}^{2} -
v_{xi}^{2})/(2(x_{f }- x_{i})) =
(0.3 m/s)^{2}/(0.036 m) = 2.5 m/s^{2}.

(b) v = at, t = v/a = (0.3 m/s)/(2.5 m/s) = 0.12 s

or
v_{x(avg) }= (v_{xf }+ v_{xi})/2 = 0.15 m/s. ∆x = v_{x(avg)}∆t,

∆t = ∆x/v_{x(avg)}
= 0.018 m/(0.15 m/s) = 0.12 s.

(c) The figure below shows a
typical electrocardiogram waveform.
0.12 s seems a reasonable acceleration time.