Si t1 et t2 désignent le débutet la

Si t1 et t2 désignent le débutet la fin de la phase de décélération, la durée de la décélération est ∆t21 = t2 − t1. On sait que durantcette phase la vitesse de la camionnette est donnée par (voir cours : mouvement rectiligne uniformémentdécéléré) : v = a(t − t1) + v1. On a alors : v2 = a(t2 − t1) + v1 = a∆t21 + v1 =⇒ ∆t21 = (v2 − v1)/a soit,∆t21 = (15 − 25)/(−2.5) = (−10)/(−2.5) = 4 s

Results (English) 1:

If t1 and t2 designate the start and end of the deceleration phase, the duration of the deceleration is ∆t21 = t2 − t1. We know that during this phase the speed of the van is given by (see lesson: uniformly decelerated rectilinear movement): v = a(t − t1) + v1. We then have: v2 = a(t2 − t1) + v1 = a∆t21 + v1 =⇒ ∆t21 = (v2 − v1)/a i.e., ∆t21 = (15 − 25)/(−2.5) = (− 10)/(−2.5) = 4 s

Results (English) 2:

If T1 and T2 indicate start
At the end of the deceleration phase, the deceleration duration is ∆ t21=t2 − t1. We know, Durant
The speed of the truck at this stage is given by (refer to course: Uniform Linear Motion)
Deceleration: v=a (t − t1)+v1. We have: V2=A (T2 − T1)+V1=A ∆ T21+V1=⇒ T21=(V2 − V1)/A
∆ T21=(15-25)/(2.5)=(-10)/(2.5)=4 seconds

Results (English) 3:

If t1 and t2 represent the beginning
And at the end of the deceleration phase, the duration of deceleration is Δ t21 = t2t1. It is reported that Durant
The speed of the truck in this phase is given (see course: uniform linear motion
Deceleration): v = a (Tt1)+v1. v2 = a(t2 − t1) + v1 = a νt21 + v1 = ⇒ ∆t21 = (v2 − v1)/a,
∆t21 = (15 − 25)/(−2.5) = (−10)/(−2.5) = 4 s