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 $8\,s$.

Sketch of the problem

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:


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


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?.


1 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$^2$.

  • $\begingroup$ Gee I totally overlooked that. When I saw this problem I initially thought that the object was at rest. But can we infer that before hand given the conditions of the problem?. From the graphic as it was shown it gave me that impresion but is it a norm?. When can we say strictly that an object is at rest?. From this graph? or from a velocity vs time?. $\endgroup$ Nov 2, 2019 at 0:03
  • $\begingroup$ @Chris Steinbeck Bell - It says the object has an initial speed 4m/s and the graph starts at zero. So, it is logic to infer that. $\endgroup$
    – Quanto
    Nov 2, 2019 at 0:55
  • $\begingroup$ Sorry. I missed that part. But its clear as you mentioned. However if no information would had been given can we say that the object is at rest? or still insuficient information to conclude that? $\endgroup$ Nov 2, 2019 at 2:53
  • $\begingroup$ @ChrisSteinbeckBell - Please let me know if the answer is clear, or acceptable. $\endgroup$
    – Quanto
    Nov 18, 2019 at 19:34

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