A tricky question on kinematics asked in a competitive exam

Question

So I found this rather ambiguous question based on kinematics from a competitive examination, the question with options is - A car travels in such a way that its velocity becomes twice the previous velocity every two seconds. If the average velocity after 6 seconds is 14 m/s, what will be the distance travelled by it in first 4 seconds?

           A)12m    B)36m     C)48m     D)56m

My reasoning was as follows: Average velocity = total distance travelled / total time

thus total distance travelled = 14*6 = 84m

Now, assuming the initial velocity is v

2v + 2(2v) + 2(4v) = 84........... (Calculating distance for every 2 second interval of 6 seconds by using v*t and equating it to 84)

solving this gives v= 6 m/s

then calculating distance travelled in 4 seconds by (vt + 2vt) = 6.2 + 12.2 = 36m

Now, I later realized that I made an assumption that the velocity of the car doubles in discontinuous steps after every 2 seconds ( like say the velocity is v for the first 2 seconds and then suddenly step ups to 2v for the next 2 second interval) rather than evolving in a continuous curve like graph.

But the question doesn't specify if the velocity evolves in a continuous exponential curve or is it expected for the candidate to assume that the velocity changes in discrete steps.

Now my genuine doubt is- had it been the case that we were supposed to evaluate the answer by considering (v-t) graph as continuous

by expressing V(t) = 2^(t/2)*v(initial)

is it still possible to calculate the distance travelled in first 4 seconds having given only the average velocity ?? OR is the question incomplete ?

Answer 1

Yes. At each point in every two second interval, the velocity will be exactly twice what it was two seconds before, regardless of how the velocity varies over the interval. Therefore the average velocity (and hence the distance travelled) over that two second interval will be exactly twice what it was in the preceding two-second interval. Expressed mathematically, \begin{align} \frac{1}{2}\int_{t_0}^{t_0+2}v(t)dt&=\int_{t_0}^{t_0+2}v(t-2)dt\\ &=\int_{t_0-2}^{t_0}v(t)dt\\ &=2\left(\frac{1}{2}\int_{t_0-2}^{t_0}v(t)dt\right)\ . \end{align} Therefore, if the car travels a distance $\ d_1\ $ in the first $2$ seconds, it will travel $\ 3d_ 1\ $ in the first $4$ seconds and $\ 7d_1\ $ in the first $6$ seconds. You're told the average velocity over the first $6$ seconds, which will be $\ \frac{7d_1}{6}\ ,$ is $\ 14$m/s. Therefore $\ d_1=12$m, and the distance travelled by the car in the first $4$ seconds is $36$m.

Answer 2

Using calculus, to treat $v$ as a continuous function.

Velocity doubles every 2 seconds: $v(t) = a2^{\frac t2}$

$a$ is the initial velocity, which at this time we do not yet know.

Distance traveled is the integral of velocity.

$x(t) = a\int_0^t e^{\frac t2\ln 2} \ dt\\ = a(\frac {\ln 2}{2})(2^{\frac t2}-1)$

If the car had an average speed over the first 6 seconds of 14 units per second. $x(6) = (6)(14) = 84$

Which we can compare with the formula we derived above to solve for the unknown initial condition.

$x(6) = a\frac {\ln 2}{2}(2^3-1) = 84$

$a\frac {\ln 2}{2} = 12$

We have a function for $x(t)$

$x(t) = 12(2^{\frac t2})$

And we can evaluate it at $t=4$

$x(4) = 12 (2^2 - 1) = 36$ units