Questions: EXERCICE 5 Une mouche M; supposée ponctuelle, est posée sur une table à coussin d'air et on enregistre ses différentes positions successives à intervalle de temps régulier τ=80 ms. On obtient ; à l'échelle 1 / 10, l'enregistrement suivant qui comporte deux phases :

EXERCICE 5 Une mouche M; supposée ponctuelle, est posée sur une table à coussin d'air et on enregistre ses différentes positions successives à intervalle de temps régulier τ=80 ms. On obtient ; à l'échelle 1 / 10, l'enregistrement suivant qui comporte deux phases :
Transcript text: EXERCICE 5 Une mouche $M$; supposée ponctuelle, est posée sur une table à coussin d'air et on enregistre ses différentes positions successives à intervalle de temps régulier $\tau=80 \mathrm{~ms}$. On obtient ; à l'échelle $1 / 10$, l'enregistrement suivant qui comporte deux phases :
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Solution

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Solution Steps

Step 1: Nature of motion between M0 and M5

Between points M0 and M5, the recorded points are equidistant and aligned. This indicates uniform rectilinear motion.

Step 2: Velocity vector at t1

The velocity vector at any instant between t0 and t5 is constant. It is oriented in the direction of motion (along the line M0M5). To calculate its magnitude, we use the formula v = d/t. The distance between M0 and M1 represents the distance covered during time τ = 80ms. Measure this distance (let's call it 'x' cm) on the paper. Considering the 1/10 scale, the real distance is 10x cm = 0.1x m. So, v = (0.1x m) / (0.08 s) = 1.25x m/s. To represent this, draw a vector along the direction of M0M5 with length 1.25x cm (since 1 cm represents 1 m/s).

Step 3: Velocities at t2 and t4

Since the motion is uniform rectilinear between M0 and M5, the velocity remains constant. Therefore, the velocities at t2 and t4 are the same as at t1, which is 1.25x m/s (calculated in the previous step and where 'x' is the measured M0M1 distance in cm).

Final Answer:

  1. The motion between M0 and M5 is uniform rectilinear motion.
  2. The velocity at t1 is directed along M0M5 with a magnitude of 1.25x m/s (x being M0M1 distance in cm on the paper). The vector is represented by an arrow of length 1.25x cm.
  3. The velocities at t2 and t4 are equal to the velocity at t1 (1.25x m/s) because of uniform motion.
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