A ball is thrown vertically upwards with an initial velocity of 39.2 m/s - Leaving Cert Applied Maths - Question 1 - 2008

To calculate the distance travelled in 5 seconds, we can use the formula:

$s = ut + \frac{1}{2}at^2$

Here, the initial velocity (u) is 39.2 m/s, the time (t) is 5 seconds, and the acceleration (a) is -9.8 m/s².

Now substituting these values:

$s = 39.2(5) + \frac{1}{2}(-9.8)(5)^2$

Calculating the terms:

$s = 196 - \frac{1}{2}(9.8)(25)$

$s = 196 - 122.5$

Therefore,

$s = 73.5 \text{ m}$

To find the acceleration of P and Q, we start with the speeds:

• The speed of Q after 2 minutes is 65.5 m/s.
• Initially, the speed of Q was 5.5 m/s.

Using the formula:

$v = u + at$

We set up for Q:

$65.5 = 5.5 + a_Q \cdot (120)$

Rearranging gives:

$a_Q = \frac{(65.5 - 5.5)}{120} = \frac{60}{120} = 0.5 \text{ m/s}^2$

For particle P, knowing that it starts at 23 m/s, we can set:

$s_P = u_P t + \frac{1}{2} a_P t^2$

The distance formula gives:

$4260 = 23(120) + \frac{1}{2} a_P (120)^2$

Solving this will give us:

$a_P = \frac{4260 - 2760}{7200} = \frac{1500}{7200} = 0.208 \text{ m/s}^2$

To find the speed of P when Q overtakes it, we note that at the moment they have equal speed:

Let the speed of P when Q overtakes be v_P:

Given:

• Initial speed of P = 23 m/s
• Q's acceleration = 0.5 m/s²,
• Time taken = 120 seconds,

Using the formula:

$v_P = u_P + at$

Substituting values:

$v_P = 23 + (0.5)(120) = 23 + 60 = 83 \text{ m/s}$

The distance P is ahead of Q can be calculated using their respective distances at equal speeds:

For Q:

$s_Q = u_Q t + \frac{1}{2} a_Q t^2 = 5.5(120) + \frac{1}{2}(0.5)(120)^2$

Calculating this:

$s_Q = 660 + \frac{1}{2}(0.5)(14400) = 660 + 3600 = 4260 \text{ m}$

For P:

$s_P = 23(120) + \frac{1}{2} a_P t^2$

Calculate P's distance:

$s_P = 2760 + (0.125)(14400) = 2760 + 900 = 3660\text{ m}$

Thus, the distance P is ahead:

$distance = s_Q - s_P = 4260 - 3660 = 600 \text{ m}$