Distance and Constant
Velocity
Problems
Since
the velocity is constant (zero acceleration),
the vector straight-line distance travelled is in the same direction as
the
velocity vector. Typically the problem has a mixture of distance and
velocity
vectors. Since diagrams must always use vectors of the same type, it is
best to
convert velocities to distances using the simple relationship 
where t is
the travel
time. The distances combine to form simple polygons. For instance if
three vectors
are involved, one has a triangle as shown below.
The
key to solving the problem is to realize
that each vector can be decomposed into its x and y
components as
shown below.
The
x and y components of the
vectors are then related by simple equations. Notice in Figure 1 that
the tail
of D3 is at the nose of D2, that
is D3
is added to D2. Since
the tail of D1 is at the tail of D2
and the
nose of D1 is at the nose of D3
we have the
vector equation D1 = D2 + D3.
This vector equation has two parts for us to work with D1x
= D2x
+ D3x and D1y = D2y
+ D3y.
Example
Two vehicles leave the same spot
at the same time. Vehicle 1 travels at 25 km/h at f = 25° west
of north. Vehicle 2 travels at 18 km/h at b = 40° west
of north. They stop at the same time and are now 200 km apart.
- How long where they travelling for?
- If vehicle 1 turns to face vehicle 2, what direction would it be looking in?
We first sketch our diagram with the given information. It looks much like Figure 1.
Our equations are
–25tsin(25°) = 18tcos(40°)t
– 200cos(a)
[1a]
25tcos(25°) = 18tsin(40°)t + 200sin(a)
[1b]
To solve we should collect terms with t on the same side
–t{25sin(25°) +
18cos(40°)}
= –200cos(a)
[2a]
t{25cos(25°) –
18sin(40°)}
= 200sin(a)
[2b]
and simplify
t{24.3543}
= 200cos(a)
[3a]
t{11.0875}
= 200sin(a)
[3b]
To solve
this we either eliminate t or a – but
we don’t have a by itself,
we have cos(a) and sin(a). In cases
like this we usually need to use one of
the trigonometric identities: tan(a) = sin(a)/cos(a), cos2(a) + sin2(a) = 1, or sin(2a) = 2 sin(a)cos(a). The first
identity is useful here. Taking the ratio
of equation [2b] to [2a]
With this value of a we use
equation [3a] or
[3b] to find t = 3.40 hours.
So vehicles travelled 3.40 hours and vehicle 1
would have
to look at 24.5°
west of north to see vehicle 2.
