A particle P of mass 2 kg is moving under the action of a constant force F newtons - Edexcel - A-Level Maths Mechanics - Question 4 - 2011 - Paper 1

To find the vector F, we first calculate the acceleration a over the time period from t = 0 to t = 5s. The change in velocity is:

$\Delta v = (7i + 10j) - (2i - 5j) = (7 - 2)i + (10 + 5)j = 5i + 15j$

The acceleration a is given by:

$a = \frac{\Delta v}{\Delta t} = \frac{(5i + 15j)}{5} = i + 3j$

Next, we apply Newton's second law, F = ma:

Given mass m = 2kg,

$F = 2(i + 3j) = 2i + 6j$

Thus, the vector F is given by:

$F = 2i + 6j$.

For P to be moving parallel to i, the y-component of the velocity must be zero:

The velocity function can be expressed as:

$v = u + at$

where:

• Initial velocity u = (2i - 5j)
• Acceleration a = (i + 3j)

This gives:

$v = (2i - 5j) + t(i + 3j)$

At time t, this results in:

$v = (2 + t)i + (-5 + 3t)j$

Setting the j-component to zero for parallel movement:

3t = 5\ t = \frac{5}{3}$$ Therefore, the value of t when P is moving parallel to i is:\n $$t = \frac{5}{3}$$.