The bicycle chases the car in front at the speed of 5 m / s. when it is 20 meters away from the car, the car starts at the acceleration of 1 m / s Can the bicycle catch up with the car

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• 1. (1 / 2) the car starts to move at an acceleration of 1 meter every second. At 20 meters behind the car, while the car starts to move, someone starts to ride a bicycle at the same time (1 / 2) the car starts to move forward at an acceleration of 1 meter per second. 20 meters behind the car, while the car starts to move, someone starts to chase the car at a constant speed of 6 meters per second. Can he catch up with the car (1 / 2) the car is driving at the speed of 18 meters per second. When the driver suddenly finds an obstacle in front of him, he brakes immediately. Assuming that the acceleration is 3 meters per second after braking, he decelerates evenly, and calculates the sliding distance of the car in 10 seconds (2 / 2)?
• 2. Motor vehicle D is forward, what is backward
• 3. When the car moves forward or backward, turn the steering wheel to the left or right, and the direction of the front and rear of the car is the same or the opposite
• 4. The forward and backward trajectory of a car with full wheels Is the track of forward and backward motion the same circle when the car is full of wheels to the left,
• 5. A maintenance team took a car along the road maintenance line, agreed to forward as positive, backward as negative, one day from the o to the end of work The route (unit: km) is + 10, - 3, + 4, + 2, - 8, + 13, - 2, + 12, + 8, + 5 (1) How far is the closing time from o? (2) If the fuel consumption per kilometer is 0.2 liter, how much is the total fuel consumption from O to the end of work? Formula
• 6. When someone goes from a to B, he moves at a speed of 2km per hour. He advances one meter, then retreats two meters, then advances three meters, retreats four meters Q: Someone goes from a to B at a speed of 2km per hour. He advances one meter and then retreats two meters. He advances three meters and retreats four meters. How far is he from a after 10 minutes
• 7. When the particle moves along a straight line, its acceleration is a = 2-3T (SI). If t = 2S, the particle is located at x = 1m, v = 2m / s, find 1, the velocity of the particle; 2. The equation of motion of a particle
• 8. A particle starts to move from rest with an acceleration of 1 m / S2. Some people say that its velocity at the end of the first second is 1 m / s and that at the end of the second is 2 m / s. therefore, its displacement in the first second is 1 m and that in the second is 2 M. do you think that's right? Why?
• 9. A particle starts to move in a straight line from its rest. In the first second, it moves at an acceleration of a = 1 m / S & sup2; in the second second, it moves at an acceleration of a '= - 1, and in the third, it moves at an acceleration of a = 1 m / S & sup2 =What is the total displacement of this particle after 100 s of repeated acceleration in the fourth second?
• 10. The motorcycle starts from standstill, runs for a distance with the acceleration of A1 = 1m / S2, then makes a constant speed movement, and then makes a constant deceleration movement with the acceleration of A2 = - 4m / S2 until it stops. The motorcycle has walked 1440m in total and lasted for 100s. The maximum speed of the motorcycle in this process is calculated
• 11. The car starts to move at an acceleration of 1 meter per second from a standstill. At 20 meters behind the car, while the car starts to move, someone starts to catch up at a constant speed of 6 meters per second on his bicycle. Can he catch up? What is the minimum distance between people and the car?
• 12. The car starts to move in a straight line at a constant acceleration of a = 1m / S & sup2; from a standstill, and the people at a distance of 25m behind the car chase the car at a constant speed of V = 6m / s, Can catch up with you? We need to calculate the process
• 13. Width L = 300m, river velocity V1 = 3m / s, ship velocity V2 = 5m / s in still water, find: (1) what is the shortest time for the boat to cross the river? What is the course of the boat? (2) what is the shortest displacement for the boat to cross the river? What is the time for the boat to cross the river?
• 14. The width of the river is d = 300 m, the velocity of the river is V1 = 1 m / s, the velocity of the ship in still water is V2 = 3 M / s, and the course of the ship forms a 30 degree angle with the upstream bank How long does it take for it to cross the river (3) How can the course reach the opposite shore (4) How to cross the river in the shortest time, (2) 320m upstream （3）cosθ=1/3 (4) The bow is always perpendicular to the other side of the river, t = 100s
• 15. When the width of the river is 420m, the speed of the ship in still water is 4m / s, and the flow speed is 3m / s, the shortest time for the ship to cross the river is? The answer is 105s, hope to have a complete process
• 16. The width of the river is 200m, the static water speed of the ship is 5m / s, the water speed is 3m / s, the shortest time to cross the river? The shortest time for the ship to cross the river? The moving distance along the downstream when it is perpendicular to the opposite bank Seeking solution urgently
• 17. River width L = 300m, water velocity V1 = 3m / s, ship velocity V2 = 5m / s in still water River width L = 300m. Water velocity V1 = 3m / s. ship velocity V2 = 5m / s in still water. Find 1 to cross the river in the shortest time. Find 2 to cross the river in the shortest displacement. Find 3 to cross the river in the shortest displacement. When the bow of the ship is 37 degrees to the upstream bank, find the sailing time to reach the opposite bank
• 18. It is known that the width of the river is 300 m, the flow velocity is 1 m / s, and the velocity of the ship in still water is 3 M / s (1) Crossing the river in the shortest time; (2) crossing the river with the least displacement; (3) Reach 100m downstream of the opposite bank; (4) reach 100m upstream of the opposite bank These four conditions are used to calculate: the degree of the angle between the ship's course and the upstream opposite bank? The time to cross the river? Let v 1 be the velocity passing through the flow direction, V 2 be the velocity perpendicular to the flow direction, and t be the time to cross the river V1²+V2²=3² （V1+1)*t=100 t=300/V2. The same as (3): V1 & sup2; + V2 & sup2; = 3 & sup2; (v1-1) * t = 100, t = 300 / V2, t =? α =? Can you explain in detail, especially the three energies 3 & sup2; in V1 & sup2; + V2 & sup2; = 3 & sup2? Thank you very much for your answer, sincerely ask!
• 19. The width of the river is d = 300 m, the velocity of the river is V1 = 1 m / s, the velocity of the ship in still water is V2 = 3 M / s, and the course of the ship forms a 30 degree angle with the upstream bank The width of the river is d = 300m, the velocity of the river is V1 = 1m / s, the velocity of the ship in still water is V2 = 3m / s, and the course of the ship forms a 30 degree angle with the upstream bank (2) Right across the bank (3) How can the course reach the opposite shore (4) What is the shortest way to cross the river
• 20. River width L = 300m, river velocity V1 = 3m / s, ship velocity V2 = 5m / s in still water If crossing the river with the shortest displacement, what is the sailing time and angle of the ship In addition, if the ship's speed is vertical, what is the combined speed of the ship's speed and water speed? Is V1 vertical to V2? What is the angle