A particle P moves along a straight line - Leaving Cert Applied Maths - Question 10 - 2019

The acceleration of P is given by the derivative of speed: $\frac{dv}{dt} = 2at + b$ Substituting a and b: $\frac{dv}{dt} = 2(-2)(4) + 0 = -16 \, \text{m/s}^2$

To calculate the distance travelled in the third second, we need to evaluate the distance at t = 3 and t = 2: Using the equation of motion: $s = \int_0^t v \: dt$ For t = 3: $s(3) = \left[ \frac{2}{3}t^3 - 10t^2 + 15t \right]_0^3$ Calculating gives: $s(3) = 18 - 30 + 45 = 33$

For t = 2: $s(2) = \left[ \frac{2}{3}(2)^3 - 10(2)^2 + 15(2) \right]_0^2$ Calculating gives: $s(2) = 10 - 40 + 30 = 0$ Thus, distance travelled in the third second: $\text{Distance} = s(3) - s(2) = 33 - 0 = 33 \, \text{m}$

Using Newton's second law, we start with: $m\frac{dv}{dt}=mg - kmv$ Integrating gives: The integral becomes: $\int \frac{dv}{g - kv} = \int dt$ We solve this to find: $\frac{1}{k} \ln |g - kv| = t + C$ Solving for v gives: $v = \frac{u - \frac{g}{k}}{(g - u)e^{-kt}}$

Using: $v = 10 - (10 - 0.2)e^{-0.98t}$ Integrating: $s = \int_0^4 v \, dt$ After calculating: $s = 10.00 - 0.20 = 39.8 \, m$