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Suppose $\vec{v}$, $\vec{u}$, $\vec{x}$, $\vec{a}$ and $t$ are final velocity, initial velocity, displacement and acceleration vectors and time.

Here are three equations of motion:

$\vec{v}$ = $\vec{u}$ + $\vec{a}$$t$

$\vec{x}$ = $\vec{u}$$t$ + $\frac 12$$\vec{a}$$t^2$

$\vec{v}^2$ = $\vec{u}^2$ + $2\vec{a}$$\vec{x}$

I wanted to know:

(1)What type of vector product is used in these equations?

(2)How can we obtain time from first equation if $\vec{v}$, $\vec{u}$,$\vec{a}$ are given as division of vectors in not allowed?

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    $\begingroup$ My guess would be that the implied vector product is the dot product, which can be used to express the magnitude of a vector $||\vec v||^2 = \vec v \cdot \vec v$. For your second question, if all the vectors are known but time is not, you can simply restrict to looking at one coordinate/one dimension of the problem and solve for t there. The equation is quadratic so there are at most two solutions, and if you need to distinguish which of the two is the right answer (if the answer is nontrivial), you can solve the same equation in the second coordinate, and compare the two solutions for t $\endgroup$
    – Rob
    Commented Sep 15, 2020 at 9:42
  • $\begingroup$ Division by a dot product, $\vec a \cdot \vec a$ is allowed, if this quantity is non-vanishing, so dotting the first equation by $\vec a$ gets you there. $\endgroup$ Commented Sep 15, 2020 at 14:48

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(1) The vector product in the third equation is the dot product, which you can see as follows:

$ |\vec v|^2 = \vec v . \vec v = (\vec u + \vec a t) . (\vec u + \vec a t) \\ = \vec u . \vec u + 2 \vec u . \vec a t + \vec a . \vec a t^2 \\ = \vec u . \vec u + 2\vec a .(\vec u t + \frac 1 2 \vec a t^2) \\ = \vec u . \vec u + 2\vec a .\vec x = |\vec u|^2 + 2\vec a .\vec x$

(2) If $\vec v - \vec u$ is a scalar multiple of $\vec a$ then you are fine. If $\vec v - \vec u$ is not a scalar multiple of $\vec a$ then there is an error in one the vectors you have been given.

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  • $\begingroup$ Please tell what is scalar multiple. $\endgroup$ Commented Sep 17, 2020 at 5:11
  • $\begingroup$ @ShreyanshKuntal A scalar multiple is “ordinary” multiplication by a real number (in a real vector space). So $(4,2)$ and $(-2,-1)$ are both scalar multiples of $(2,1)$ (and of each other) but $(2,0)$ and $(3,2)$ are not. $\endgroup$
    – gandalf61
    Commented Sep 17, 2020 at 10:27

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