Question

A bird flies for 4s with a velocity of $|t - 2|m/s$ in a straight line, where t is time in seconds. It covers a distance of
A. 2m
B. 4m
C. 6m
D. 8m

Answer 1

Hint: Distance covered can be calculated by taking the product of time taken by the bird and the velocity with which it is moving. Velocity is equal to the time derivative of distance travelled.

Complete Step-by-Step solution:
$ t = 2s $
$v = |t - 2|m/s $
Velocity is a function of time here and we know that distance is equal to the product of time and velocity and must be a positive quantity. Therefore, for first 2sec,
$v = (2 - t)m/s$ and after 2 sec, $v = (t - 2)m/s$. Now, velocity is the time-derivative of distance. In differential form,
$ v = \dfrac{{dx}}{{dt}} \\
 \Rightarrow dx = vdt \\ $
Integrating both sides, we get
$ \int {dx = \int {vdt} } \\
   \Rightarrow x = \int\limits_0^2 {(2 - t)dt} + \int\limits_2^4 {(t - 2)dt} \\
   \Rightarrow x = \left[ {2t} \right]_0^2 - \left[ {\dfrac{{{t^2}}}{2}} \right]_0^2 + \left[ {\dfrac{{{t^2}}}{2}} \right]_2^4 - \left[ {2t} \right]_2^4 \\
   \Rightarrow x = 4 - 2 + 8 - 2 - 8 + 4 \\
   \Rightarrow x = 4m \\ $
Hence, the correct answer is option B.

Note: If we directly substitute the value of time t into the expression for velocity and then multiply velocity with time to get distance then we obtain option D which is wrong. This is because the velocity being a function of time means that there is instantaneous change taking place with time and we need to calculate the amount of change at every point which is done by integrating as done in the above solution.