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
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
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
- After the stop:
S = 115km
t = 4h-1h-1h20min = 1h40min = 1.66h (using simple three-rule)
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:
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?
S = 10km
v = 70km / h
S = 70t
10 = 70t
0.14h = t
t = 8.57min (using simple three rule)
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 = 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.
v = 50km / h
S = 10km