Senior 1 Physics Catch up Questions Please answer in detail, thank you 30 16 17 15

If it catches up after t seconds, then.

t*1=5t=2t^2-40t+100

2t^2-45t+100=0

t1=,t2=20

Because t>10

So t=20s

The maximum distance is.

B's velocity is equal to A's.

Namely. t = when A moves the second is the distance is maximum.

The distance is at this point. Rice.

B sets off "time t" to catch up with A.

A displacement: x1=v1*(t+10).

B displacement: x2 = (1 2)a*t 2

Let x1=x2

v1*(t+10)=(1/2)a*t^2

t+10)=

solve, t=10s

Let x=x1-x2=

v1*(t+10)}-

t+ This is a "quadratic function", using the method of finding the "extremum", we get:

xmax=start, A's speed is greater than B's, and the distance between the two cars increases; Later, B's speed was greater than A's, and the distance between the two cars decreased.

The distance between them is greatest only when the velocity is equal. -- That's the way it goes.

1.A few seconds after B sets off to catch up with A, set t, s A = 10 + ts B =

s A = s B.

t=10s2.B catches up with A and the maximum distance between them? Let ts=sA-sB=10+<=

When B catches up with A, the displacement of both is equal.

Rule. sA = v A*(t B + t) 1

s B = 1 2 * at B 2 2

Again. sA = sB3

Lianli 1-3t B = 10s

The maximum displacement between A and B occurs when the velocity of A and B is equal, so.

t max = v A 4

s max = v A * (t + t max) -1 2 * at 2 5 simultaneous s max =

1Set the time to x

x=Solve the system of equations, sorry I can't type the root number with the computer.

2.When the velocity of A and B is equal, the distance is maximum. for meters.

A*t=10

t = 10v A = v B.

The distance is maximum. Therefore s=10+(

Column: V A(t+10) = 1 2 at 2

It can be solved t=-5 or 10 because it is a practical problem, t should be taken for 10 seconds.

Let the distance between A and B be x

x=vA(t+10)- 1 2 at 2

1/5(t^2 - 5t -50)

1 5 (t - 5 2) 2 + 45 4 so B starts seconds after their spacing is the largest, which is meters.

3 minutes just catch up, let the acceleration time t, then the time of constant speed is 180s-t, the distance traveled by the acceleration time is (0+30)t 2=15t, and the distance traveled in the uniform time is 30 (180-t).

There is 15t + 30 (180-t) = 1000 + 20 * 180 to get t = 160 3 s

a=v/t=30/(160/3) =

Units are ignored.

Let the electric car catch up in exactly three minutes, and the acceleration will be a

Then 1 2*[a*(30 a) squared]+30*(3*60-30 a)=20*3*60+1000

Find a=9 16

The magnitude of the train's acceleration:

a=v1/t=(20/40)m/s^2=

t1=v2-v1/(-a)=6-20/(

Displacement of the train:

s1=v2 2-v1 2 (-2a)=6 2-20 2 (-2*displacement of the wagon during this period:

s2=v2t1=6*28m=168m

Because when the train found the wagon, the distance between the two cars was l=180m

And because: S1 = 364m> S2 + L = 168m + 180m = 348m will collide with the two vehicles.

a=20/40=

When the speed of the two cars is equal.

t=△v/a=28s

S cargo = VT = 168M

S1 = v average t = 13 * 28 = 364 m

S1-S cargo = 196m will collide.

1. It is known that a car with good braking performance can be braked at a distance of 56 meters when driving at a speed of 80 kilometers per hour; At a speed of 48 km/h, it can be braked at a distance of 24 meters. Assuming that for both rates, the driver is allowed to react in the same amount as the acceleration when braking, what is the allowable driver reaction time?

Substitute the formula to solve the system of equations.

2. A and B two cars at the same rate v0 in the horizontal plane to do a uniform linear motion, at a certain time B car first with the size of a acceleration to do a uniform deceleration motion, when the rate is reduced to zero, a car also with a size of a acceleration to do a uniform deceleration motion, in order to avoid collision, in the B car begins to do a uniform deceleration motion, a B two cars should be at least how much distance?

First calculate the displacement of vehicle B in the process of deceleration, and then calculate the displacement of vehicle A in the process of vehicle B deceleration, and subtract them.

3. The car is moving forward on a straight highway at a speed of 10 meters per second, and suddenly finds that there is a bicycle in front of it at a speed of 4 meters per second to do a uniform linear motion in the same direction, and the car closes the throttle to do a uniform deceleration movement of 6 meters per square second.

The critical state is that when the car slows down to 0, it just collides with the bicycle (relative displacement = 0) The specific method is the same as 2

4. A and B two cars on the same straight one-way road B before and after the same townspeople drive at a constant speed, A and B two cars speed respectively 40 meters per second and 20 meters per second, when the two cars are close to 250 meters away from the two cars brake at the same time, it is known that the braking acceleration of the two cars is meters per square second and 1 3 meters per square second, ask car A will hit car B?

Calculate the displacement of the two cars in the process of deceleration When (the displacement of car B + 250 meters) is less than the displacement of car A, it will collide.

5. When a person is 45 meters away from the right entrance of the tunnel in a 100-meter-long mountain tunnel, he suddenly finds a rapid train (with a speed of 72 kilometers per hour) appearing 200 meters away from the right entrance of the tunnel, and the person has no time to think, so he runs along the tunnel to the left exit (at a speed of 4 meters per second) (1) Is the person choosing the right direction (safer than choosing the other direction)? (2) If the person happens to be running in a safe way when he finds the train, how far should this position be from the right exit of the tunnel?

Choose the right one. 2 Let the distance between the person and the right end be x, then if the person runs to the left, 100-x) 4 <= (200+100) 20=15

So x>=40m x 4 < = 200 20 = 10 x < = 40m if the person runs to the right

Therefore, if it is safe for people to run on both sides, they will take the intersection of these two ranges, so this distance is 40m away from the hole on the right

1.The train in front is used as a reference, there are s=120m, v0=14m s, a=t=v0 a=

s'=1/2v0*t=7*>120m

It is known that there will be a collision.

1) When one accelerates and one decelerates, it can be known that the maximum distance is the same speed of two cars, set the time t and the speed as v'

There are 10-4*t=v',1*t=v', and the solution gives t=2s,v'=2m s car travel distance s1 = 10*

The distance traveled by car B is s2=

The maximum distance is s=s1-s2=10m

2) Let the time of re-encounter be t, s1=10*

s2=s1=s2 gives t=4s

Taking an oncoming car as a reference, the initial velocity of the truck is 134 km h, which is 40 m s, and the acceleration is 15 m s 2

The reaction time is that the time to decelerate to 0 is t=40 15=8 3 and the driving distance is s=40*

It can be seen that the two cars will collide, and the time from the deceleration of the reaction to the collision time t, then 40 * solution t=

It can be seen that when the car collides, it is stopped, and the time to slow down to the notification is 15 and the driving distance is 15*

It can be seen that the distance traveled by the truck is.

From 25*, we get t=

v = At the time of collision, the truck speed is.

To do this you need to clarify the nature of the motion of the two objects, and the other is the relationship between the two (e.g. displacement, time, velocity, acceleration). If you want to determine the connection between the two, you must do a good analysis, such as the first one: if the passenger car does not do anything, it will definitely collide, because the V passenger "V train", but the brake may not allow you to avoid the collision, because there is still a V passenger "V train during the deceleration process.

Then when V passenger = V train collides or not, the displacement relationship between the two of them: whether X passenger is greater than X train + 120, the time of motion is equal: T passenger = T train.

If x customers hope to help you!!

2. When the speed of the two cars is equal, the distance is the farthest, and the solution is obtained by the equation 0+1*t=10-4*t, which is equal to the speed

s A = 10 * 2 - 1 2) * 4 * 2 2 = 12 m s B = (1 2) * 1 * 2 2 = 2m

So the maximum distance = 12-2 = 10m

When they meet again, the displacement of the two cars is the same 10*t - 1 2)*4*t 2=(1 2)*1*t 2, and the solution is t=4s

Question 1: When the speed of the two buses is equal, the distance between the two is the closest, just calculate the distance between the two when the passenger car slows down to 6 meters per second.

Analysis: The initial velocity of the bomb is the same as that of the airplane, the horizontal velocity is v1 = 100m s, the vertical velocity is 0, the bomb hits the car from the launch to the ground, the time taken is t = (2h g) = (2 * 500 10) = 10s, the horizontal movement distance is s1 = v1 * t = 100 * 10 = 1000m, and the car travel distance in the same time is s2 = v2 * t = 20 * 10 = 200m, so the plane should be in the opposite direction of the car distance car s =S1-S2 = 1000-200 = 800m to drop bombs.

Drop a bomb at a distance of 400 meters,

Solution: Vertical direction: 1 2gt2=500

Solution: t=10s

Horizontal: s = (100-20) * 10 = 800m, so the plane should drop bombs at a distance of 800m from the car horizontally.

The bomb flight time t=10s, the horizontal flight distance of the bomb is 1000m, and the driving distance of the car during this time is 200m, so the plane should be dropped 800m behind the car.