Question

. The velocity-time graph of a passenger lift is shown in the fig. below. (note the time corresponding to pt. A is 4.6s. Answer the following questions, A. Mention the kind of motion of the lift represented by parts OA and BC of the graph. B. What is the acceleration of the lift- i) during the first two seconds? ii) between the 2nd and 10th second? iii) during the last two seconds?

Answer 1

Answer:

A) accelerative

B) i) 2,3 $m/s^{2}$

ii) 0  $m/s^{2}$

iii) -2,3  $m/s^{2}$

Explanation:

A) The beginning and ending part of the graph are not parallel with the x axis, therefore, the velocity is not the same, meaning that during those periods the lift is accelerating/decelerating.

B) i) The formula is v=at, where v is velocity, a acceleration and t time. From there we can see that a=v/t. According to the text of the task, v=4,6 m/s. The time equals t=2 s and now we have a=v/t = (4,6/2) $m/s^{2}$ = 2,3 $m/s^{2}$

ii) This time the motion is uniform, meaning the a=0 $m/s^{2}$.

iii) Analogous to the t) task, the time is the same, as well as the velocity, but now the lift is decelerating so we just put the negative sign and have the solution  -2,3 $m/s^{2}$.