Find the values of t for which the velocity of the car is parallel to the vector (i + j)
v = (3r^2 - 2t + 8)i + (5t + 6)j ms^-1
Its doing my head in not being able to work this out
Find the values of t for which the velocity of the car is parallel to the vector (i + j)
v = (3r^2 - 2t + 8)i + (5t + 6)j ms^-1
Its doing my head in not being able to work this out
Hello, Carl!
Find the values of $\displaystyle t$ for which the velocity of the car
is parallel to the vector $\displaystyle \vec i + \vec j$
. . $\displaystyle \vec v \;= \;(3t^2 - 2t + 8)\vec i + (5t + 6)\vec j$
A vector parallel to $\displaystyle \vec i + \vec j$ has the form: .$\displaystyle \vec v \:=\:a\vec i + a\vec j$
. . That is, the coefficients are equal.
So we have: .$\displaystyle 3t^2 - 2t + 8 \;=\;5t + 6 \quad\Rightarrow\quad 3t^2 - 7t + 2 \:=\:0$
Factor: .$\displaystyle (t - 2)(3t - 1) \:=\:0 \quad\Rightarrow\quad\boxed{ t \;=\;2,\:\frac{1}{3}}$
Another way may be to use dot products.
Recall that
$\displaystyle \mathbf{a}.\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos{\theta}$.
Since the vectors will be parallel, the angle between them is 0, and so $\displaystyle \cos{\theta}=\cos{0} = 1$.
So in other words, in this case...
$\displaystyle \mathbf{a}.\mathbf{b} = |\mathbf{a}||\mathbf{b}|$
So evaluate the dot product, evaluate the product of the moduli, and solve for t.