Two bodies depart with 15 s of difference from the same point in the same direction. Knowing that the speed of the first one is 72 km/h, what should be the speed of the second to reach it in 90 s?
Note: You can assume that both bodies move with uniform rectilinear motion.
- Speed of the first body: V1 = 72 km/h = 20 m/s
- Difference in time of departure between the two bodies ∆t = 15 s
- Instant in which the bodies meet: t = 90 s
- Two bodies begin their motion at different times. The time difference between them is 15 s. Therefore, when the first body has been in motion for 90 s, the second one has only been 90 - 15 = 75 s
- For them to meet after 90 s, the distance traveled by the second body in 75 s must be the same as the distance the first one travels in 90 s
- Keep in mind that we can use the sign convention in rectilinear motion that allows us to use scalar magnitudes instead of vectors to describe the motion
The expression that allows us to determine the position of each body as a function of the speed and of the time is:
If t1 is the time that the first body is in motion, then for the first body we get:
If t2 is the time that the second body is in motion, then for the second body we get:
Both bodies depart from the same point, therefore x01 = x02 = 0 , but they do so at different times: t2 = t1 - 15 s. . When they meet, they are at t1 = 90 s and t2 = 90 - 15 = 75 s. Additionally, when they are in the same position x1= x2, resulting in:
Ambos cuerpos parten del mismo punto, por tanto x01 = x02 = 0 , pero lo hacen en instante de tiempos distintos: t2 = t1 - 15 s. En el momento en que se encuentran t1 = 90 s y t2 = 90 - 15 = 75 s. Además, cuando se encuentran están en la misma posición, es decir, x1 = x2 , quedando: