III.3. Recapitulative solved problems- Movement and Rest
🔓 High difficulty recapitulative solved problems - Movement and Rest.
1. A mobile moves evenly slow to the stop with an acceleration of 1,5 m/s2, in one minute. How far did the mobile travel?
Solution:
We write down the problem data:
a = 1,5 m/s2 = constant
v0 > 0
v = 0 (mobile stops)
Δt = 1 min = 60 s
We write the acceleration formula and calculate the initial velocity, v0, of the mobile:
From the expression of the average speed we obtain:
Since in uniformly varied motion velocity is a linear function of time (i.e. constantly increasing / decreasing in the same time interval), the average velocity is the arithmetic average of the initial (v0) and final (v) velocities, we obtain:
2. Determining the uniformly varied law of motion.
Solution:
From the expression of the average speed we obtain:
Since in uniformly varied motion velocity is a linear function of time (i.e. constantly increasing / decreasing in the same time interval), the average velocity is the arithmetic average of the initial (v0) and final (v) velocities, we obtain:
From the acceleration formula we subtract the speed and then replace it in the position formula:
3. A car is moving at a constant speed of 25 m / s. A motorcyclist starts (starts) when the car passes him, with a uniformly accelerated movement, reaching a speed of 25 m / s in 10 s without stopping to accelerate. Determine the time after which the motorcyclist catches up with the car.
Solution:
We write down the data of the problem those related to the car with index 1 and those related to the motorcyclist with index 2:
v1 = 25 m/s = constant
v2 = 25 m/s = increases constantly in equal time intervals
a1 = 0 m/s2
a2 = constant
Δt2 = 10 s
For the 2 mobiles we have x0, v0, t0 = 0, Δt = t
We write the law of motion of mobile 1 (the car):
We calculate the acceleration of mobile 2 (the motorcycle) and write the law of motion:
We put the condition that the two mobiles meet: x1 = x2 and we find the meeting time:
