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202210201704 Solution to 1971-CE-AMATH-II-XX

A particle is projected vertically from the ground so that its velocity (in \mathrm{ft/sec}) in the first 4 seconds is given by

V_1=128-32t\qquad (0\leqslant t\leqslant 4)

and after the fourth second by

V_2=3t^2-32t+80\qquad (4\leqslant t)

where t is the time (in seconds) after projection.

i. Show that V_1=V_2 when t=4.
ii. Calculate the height of the particle when t=4.
iii. Calculate the height of the particle when t=5.
iv. Find the value of t when the acceleration of the particle is 40\,\mathrm{ft/sec^2}.

Roughwork.

i.

When t=4:

\begin{aligned} V_1 & = 128-32(4) \\ & = 0 \\ V_2 & = 3(4)^2-32(4)+80 \\ & = 0 \\ V_1 & = V_2 \\ \end{aligned}

ii.

\begin{aligned} h(t\in [0,4]) & = \int_{0}^{4}V_1\,\mathrm{d}t \\ & = \int_{0}^{4}(128-32t)\,\mathrm{d}t \\ & = \bigg[ 128\bigg(\frac{t^{(0)+1}}{(0)+1}\bigg) - 32\bigg(\frac{t^{(1)+1}}{(1)+1}\bigg)\bigg]\bigg|_{0}^{4} \\ & = [128(4)-16(4)^2]-[128(0)-16(0)^2] \\ & = 256\,\mathrm{ft} \\ \end{aligned}

iii.

\begin{aligned} h(t\in [0,5]) & = h(t\in[0,4]) + h(t\in[4,5]) \\ & = 256\,\mathrm{ft} + \int_{4}^{5}V_2\,\mathrm{d}t \\ & = 256\,\mathrm{ft} + \int_{4}^{5}(3t^2-32t+80)\,\mathrm{d}t \\ & = 256\,\mathrm{ft} + \big[ t^3-16t^2+80t\big]\big|_{4}^{5} \\ & = 256\,\mathrm{ft} -3\,\mathrm{ft} \\ & = 253\,\mathrm{ft} \\ \end{aligned}

iv.

\begin{aligned} a(t\in [0,4]) & = \frac{\mathrm{d}}{\mathrm{d}t}V_1=-32 \\ a(t\in [4,T]) & = \frac{\mathrm{d}}{\mathrm{d}t}V_2=6t-32 \\ \end{aligned}

\begin{aligned} 6t-32 & = 40 \\ t & = 12\,\mathrm{s}\\ \end{aligned}