## Velocity:

**1.** A monkey that jumps from branch to branch in a zoo takes 6 seconds to cross its 12-meter cage. What is his average speed?

*S = 12m*

*t = 6s*

*v =?*

**2. **A car travels from city A to city B, 200km away. It takes 4 hours, because after one hour of travel, the left front tire punctured and needed to be replaced, taking 1 hour and 20 minutes of total time spent. What was the average speed the car developed during the trip?

*S = 200km*

*t = 4h*

*v =?*

*Even though the car has been stationary for some time during the trip, the average speed is not taken into account.*

**3.** In the previous exercise, what was the speed at intervals before and after the tire punctured? Knowing that the incident occurred when 115 km were left to reach city B.

*Before the stop:*

*S = 200-115 = 85km*

*t = 1 hour *

*v =?*

*After the stop:*

*S = 115km*

*t = 4h-1h-1h20min = 1h40min = 1.66h* *(using simple three-rule)*

*v =?*

**4. **A baseball is thrown at a speed of 108m / s, and it takes 0.6 seconds to reach the batter. Assuming the ball moves at a constant speed. How far is the pitcher from the hitter?

, *if we isolate S:*

**5.** During a 100 meter race, a competitor travels at an average speed of 5m / s. How long does it take to complete the route?

*, if we isolate t*:

## Uniform Movement:

**1. **A car moves along a straight path described by the function S = 20 + 5t (in SI). Determine:

(a) the starting position;

(b) the speed;

(c) the position at time 4s;

(d) the space traveled after 8s;

(e) the instant the car passes the 80m position;

(f) the instant the car passes the 20m position.

* Comparing with the default function: *

*(a) Starting position = 20m*

*(b) Speed = 5m / s*

*(c) S = 20 + 5t*

*S = 20 + 5.4*

*S = 40m*

*(d) S = 20 + 5.8*

*S = 60m*

*(e) 80 = 20 + 5t*

*80-20 = 5t*

*60 = 5t*

*12s = t*

*(f)* *20 = 20 + 5t*

*20-20 = 5t*

*t = 0*

**2. **On a 10km slope, the maximum allowed speed is 70km / h. Suppose a car starts this stretch at a speed equal to the maximum allowable, while a bicycle does it at a speed of 30km / h. How far is the car from the bike when the car is completed?

*Car:*

*S = 10km*

*v = 70km / h*

*t =?*

*S = 70t*

*10 = 70t*

*0.14h = t *

*t = 8.57min (using simple three rule)*

*Bike*

*The time used to calculate the distance reached by the bicycle is the time the car reached the end of the journey: t = 0.14h*

*v = 30km / h*

*t = 0.14h*

*S =?*

*S = 0 + 30. (0.14)*

*S = 4.28km*

**3.** The following graph shows the time-dependent positions of two buses. One part of city A towards city B, the other from city B to city A. Distances are measured from city A. How far will the buses be?

*In order to make this calculation, we need to know the speed of either bus, and then calculate the distance traveled until the moment the two meet, where the paths intersect.*

*Calculating bus speed from city A to city B (blue line)*

*Knowing the speed, it is possible to calculate the position of the encounter when t = 3h.*

**4.** A car travels at a speed of 20m / s at first, then moves at a speed of 40m / s, as shown in the chart below. How far was the car?

*Having the graph of v *x* t, the displacement is equal to the area under the velocity line. So:*

S = Area A + Area B

S = 205 + 40(15-5)

S = 100 + 400

S = 500m

**5. **Two trains depart simultaneously from the same place and travel the same straight line with speeds of 300 km / h and 250 km / h respectively. There is communication between the two trains if the distance between them does not exceed 10km. How long after leaving will trains lose radio communication?

*For this calculation the relative speed between the trains is established, thus the motion can be calculated as if the fastest train was moving at a speed of 50km / h (300km / h-250km / h) and the other stopped.*

*Like this: *

*v = 50km / h*

*S = 10km*

*t =?*