How can I find the speed and the displacement an object has attained from acceleration graph?

Question

The problem is as follows:

An ojbect is moving along the $\textrm{x-axis}$ with an initial speed of $+4\hat{i}\,\frac{m}{s}$, its acceleration against time is given in the graph from below. Find the speed on $\frac{m}{s}$ of the object for $t=9\,s$ and its displacement (on $m$) during the first $8\,s$.

The given alternatives are:

$\begin{array}{ll} 1.&+12\hat{i}\,\frac{m}{s}\,;52\,m\\ 2.&+12\hat{i}\,\frac{m}{s}\,;41\,m\\ 3.&+14\hat{i}\,\frac{m}{s}\,;41\,m\\ 4.&+16\hat{i}\,\frac{m}{s}\,;52\,m\\ \end{array}$

This problem looks trivial but I'm stuck with the second part which is to find the displacement as for me it doesn't look very obvious. Can somebody help me with this?

What I did for the first part was to use this formula:

$v_{f}=v_{o}+at$

Since the initial speed is given and the acceleration is $a=2$ then:

$v_{f}=4+(2)(4)=12\,\frac{m}{s}$

But the displacement is where I'm stuck at:

Wouldn't it be:

$v_{f}^2=v_{o}^2+2a\Delta x$

$12^2=4^2+2(2)\Delta x$

$\Delta x = \frac{144 - 16}{4}=\frac{128}{4}=32 m$

But this does not check with any of the alternatives given. What should I do?.

Answer 1

Note that the object travels with constant speed 4m/s for the first 5s and accelerates with 2m/s$^2$ afterwards.

The speed at 9s is,

$$v_9 = v_5+ a (9-5) = 4 + 2\times 4=12m/s$$

The displacement at 5s is,

$$s_5 = v_0 (5 -0) = 4\times 5 = 20m/s$$

and the displacement at 8s is,

$$s_8 = s_5 + v_5 (8-5) + \frac12 a (8-5)^2= 20+ 12+ 9= 41m$$

Thus, the answer is (2).

You need to calculate the displacement in two segments, the first 5s with constant speed 4m/s and then the next 3s with initial speed 4m/s and acceleration 2m/s$^2$.