Question

un cycliste se déplace à une vitesse de 8,0 m/s. il prend 3,0 s à descendre une pente en accélérant à un rythme de 5,0 m/s². au bas de la pente, il freine pendant 0,5 s avant d’entrer dans une courbe à une vitesse de 18,0 m/s. quelle a été son accélération moyenne pendant sa période de freinage ?

Explanation:

Step 1: Identifier les données

Initial velocity \( u = 8.0 \, \text{m/s} \), final velocity \( v = 18.0 \, \text{m/s} \), time of acceleration \( t_1 = 3.0 \, \text{s} \), time before braking \( t_2 = 0.5 \, \text{s} \), total time for the process (acceleration + delay + braking? Wait, no, the braking period is what we need? Wait, the problem is: A cyclist moves at 8.0 m/s. It takes 3.0 s to go down a slope, accelerating at 5.0 m/s². At the bottom of the slope, it brakes for 0.5 s before entering a curve at 18.0 m/s. What was its average acceleration during braking? Wait, maybe I misread. Wait, let's re-express:

Wait, initial velocity \( u \) (before braking): when going down the slope, initial velocity \( u_0 = 8.0 \, \text{m/s} \), acceleration \( a_1 = 5.0 \, \text{m/s}^2 \), time \( t_1 = 3.0 \, \text{s} \). So velocity at the bottom of the slope (before braking) is \( v_1 = u_0 + a_1 t_1 \). Then, it brakes for \( t_2 = 0.5 \, \text{s} \), and final velocity \( v_2 = 18.0 \, \text{m/s} \). We need to find the average acceleration \( a_2 \) during braking.

Step 2: Calculer \( v_1 \)

Using \( v = u + at \), so \( v_1 = 8.0 + 5.0 \times 3.0 = 8.0 + 15.0 = 23.0 \, \text{m/s} \).

Step 3: Calculer l'accélération de freinage

Now, during braking, initial velocity \( u_2 = v_1 = 23.0 \, \text{m/s} \), final velocity \( v_2 = 18.0 \, \text{m/s} \), time \( t_2 = 0.5 \, \text{s} \).

Acceleration \( a_2 = \frac{v_2 - u_2}{t_2} = \frac{18.0 - 23.0}{0.5} = \frac{-5.0}{0.5} = -10.0 \, \text{m/s}^2 \). The negative sign indicates deceleration (braking).

Wait, but let's check the problem again. Maybe the initial velocity before braking is not from the slope? Wait, the problem says: "Il prend 3,0 s à descendre une pente en accélérant à un rythme de 5,0 m/s². Au bas de la pente, il freine pendant 0,5 s avant d’entrer dans une courbe à une vitesse de 18,0 m/s. Quelle a été son accélération moyenne pendant sa période de freinage ?"

So:

1. Initial velocity (before slope): \( u = 8.0 \, \text{m/s} \)
2. Slope: time \( t_1 = 3.0 \, \text{s} \), acceleration \( a_1 = 5.0 \, \text{m/s}^2 \)
3. Velocity at bottom (before braking): \( v_1 = u + a_1 t_1 = 8 + 5*3 = 23 \, \text{m/s} \)
4. Braking time \( t_2 = 0.5 \, \text{s} \), final velocity \( v_2 = 18 \, \text{m/s} \)
5. Acceleration \( a = \frac{v_2 - v_1}{t_2} = \frac{18 - 23}{0.5} = -10 \, \text{m/s}^2 \)

So the average acceleration during braking is \( -10.0 \, \text{m/s}^2 \) (or 10.0 m/s² deceleration).

Answer:

L'accélération moyenne pendant le freinage est \(\boxed{-10.0 \, \text{m/s}^2}\) (ou 10,0 m/s² de décélération).