# 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/s ^{2}, in one minute. How far did the mobile travel?**

**Solution:**

*We write down the problem data:*

a = 1,5 m/s^{2} = constant

v_{0} > 0

v = 0 (mobile stops)

Δt = 1 min = 60 s

*We write the acceleration formula and calculate the initial velocity, v _{0}, 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 (v _{0}) 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 (v _{0}) 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:*

v_{1} = 25 m/s = constant

v_{2} = 25 m/s = increases constantly in equal time intervals

a_{1} = 0 m/s^{2}

a_{2} = constant

Δt_{2} = 10 s

For the 2 mobiles we have x_{0}, v_{0}, t_{0} = 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: x _{1} = x_{2} and we find the meeting time:*