M1

hey can anyone help me witht the steps taken to solve this question:

1) A particle p moves with constant acceleration (2i-3j) ms^-2 . At time t seconds , its velocity is v. When t=0, v=(-2i+7j).

find the value of t when p is moving parrelel to i ??

the answer is t= 7/3

thanks

Reply 1

steve10

dv/dt = (2i-3j)

integrating,

v = (2i - 3j)t + C -------------------------(1)

at t = 0, v = -2i + 7j, therefore

-2i + 7j = 0 + C
C = -2i + 7j

substituting for C into (1)

v = (2i - 3j)t -2i + 7j
v = (2t - 2)i + (7 - 3t)j

The velocity in the i-dirn is v_i = 2t - 2
The velocity in the j-dirn is v_j = 7 - 3t

When p is moving parallel to i, then that means that it has no velocity in the j-dirn. i.e. v_j = 0
Therfore, 7 - 3t = 0
t = 7/3
=====

Reply 2

Theorist

You don't even need to use differentiation, especially given that this is M1, so it is not expected of you (at least for Edexcel anyway).

The way I did it was to use

Unparseable latex formula:

\begin{small}\mathbf{v} = \mathbf{u} + \mathbf{a}t.\end{small}

Substituting in the appropriate values,
[indent]

v(t) = \left(-2\mathbf{i} + 7\mathbf{j}\right) + t\left(2\mathbf{i}-3\mathbf{j}\right)

[/indent]

When P is travelling parallel to

Unparseable latex formula:

\begin{small}\mathbf{i},\end{small}

Unparseable latex formula:

\begin{small}\mathbf{v_{j}} = 0.\end{small}

Equating

Unparseable latex formula:

\begin{small}\mathbf{j}\end{small}

coefficients, we find from the original equation that

Unparseable latex formula:

\begin{small}7-3t = 0\end{small}

and therefore

Unparseable latex formula:

\begin{small}t = \frac{7}{3}.\end{small}

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