A car travels in a straight line from A to B, a distance of ${12}$ km, taking 552 seconds. The car starts from rest at A and accelerates for ${T_1}$ s at 0.3 $m s^{-2}$ , reaching a speed of V $m s^{−1}$ . The car then continues to move at V $ ms^{−1}$ for ${T_2}$ s. It then decelerates for ${T_3}$ s at 1 $m s^{−2}$ , coming to rest at B.

(i) Sketch the velocity-time graph for the motion and express ${T_1}$ and ${T_3}$ in terms of V.

(ii) Express the total distance travelled in terms of V and show that $13V^2 − 3312V + 72 000 = 0$. Hence find the value of V.

That's the whole question. The problem is at (ii). The quadratic has 2 solutions which are 24 and about 230.8 but the answers says it's only 24. Why is that? Is there some hidden situation in the question that excludes the 230.8? I believe 230.8 is logic and correct. Does anyone notice something I didin't? ...help please. :D

by the way the previous parts of the question are all answered via the velocity time graph and knowing that the area under graph is the distance travelled. Sadly I have no time to elaborate that.


First, from a physical point of view, 231 $m\cdot s^{-1}$ does not make sense for a car: it's insanely fast.

Secondly, the car must cover 12km in 552s. In the fastest case, $T_1=T_3=0$ and it travels all the time at $v_2$. But if $v_2\simeq 231 m\cdot s^{-1}$, covering 12 km would take at most $52$ seconds...

The issue here, I reckon, is that to arrive to the quaratic assumption one must use all the hypotheses of the problem, but these hypotheses are not all captured by the final equation (i.e., the equation is not equivalent to the problem, just implied by it). From what I can infer, it in particular does not enforce $T_1,T_3\geq 0$.

  • $\begingroup$ well the first sentence is enough for me. tried to convert the 231 to kmph and it's 828 kmph which is really so unusual. I just have some doubts upon the second sentence. From the velocity time graph. $T_1=V/0.3$ actually. and $T_2=V$. If you assume that they're equal to zero then $231*552$ must have been equal to 12000 which is not, even if you use the exact values. ...still the first sentence is enough. thanks :D $\endgroup$ – Mina Michael May 12 '14 at 12:41
  • $\begingroup$ $T_3=V$ sorry $\endgroup$ – Mina Michael May 12 '14 at 14:23
  • $\begingroup$ The second paragraph is saying that $v$ cannot be too big -- otherwise, there is no way to cover only 12km in 552s (or, as you put it, $v\cdot 552 \gg 12000$). $\endgroup$ – Clement C. May 12 '14 at 19:05

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