The car starts from the station and runs at a speed of 4m / s along the straight road at a uniform speed. After 2S, a motorcycle starts from the same station and accelerates to catch up The car starts from the station and runs at a speed of 4 m / s along the straight road at a constant speed. After 2 s, a motorcycle starts from the same station and accelerates to catch up with the car with an acceleration of 3 M / S2. How long does it take for the motorcycle to catch up with the car after it starts? How far is it from the starting point when the motorcycle catches up with the car?

The car starts from the station and runs at a speed of 4m / s along the straight road at a uniform speed. After 2S, a motorcycle starts from the same station and accelerates to catch up The car starts from the station and runs at a speed of 4 m / s along the straight road at a constant speed. After 2 s, a motorcycle starts from the same station and accelerates to catch up with the car with an acceleration of 3 M / S2. How long does it take for the motorcycle to catch up with the car after it starts? How far is it from the starting point when the motorcycle catches up with the car?

x1=v1(t+2)=4(t+2)
x2=1/2at^2=1/2*3*t^2=1.5t^2
x1=x2
4t+8=1.5t^2
3t^2-8t-16=0
(3t+4)(t-4)=0
T = - 4 / 3 rounding off
t=4s
x1=x2=4*(4+2)=24m

When a car runs at a constant speed on a straight road, the speed is V0 = 5m / s, and the acceleration of the car is a = - 0.4m/s after closing the throttle
(1) After closing the throttle, the vehicle displacement x = 30m, the time T1
(2) The distance of taxiing within 20s after closing the throttle

（1）x=v0*t+0.5at²
30m=5t-0.5*0.4*t²
∴t=10s.
（2）∵vt=v0+at
0=5-0.4t
t=12.5s
The car stopped at 12.5 seconds
∴vt²-v0²=2ax
x=31.25m.

The bus starts from the station and runs along the straight road at a speed of 4m / s. two seconds later, a motorcycle starts from the same station and accelerates to catch up with the bus at an acceleration of 2m / S2? (2) How far is it from the starting point when the motorcycle catches up with the car? (3) What is the maximum distance between the motorcycle and the car?

(1) Let V 0 (T + 2) = 12At 2. The solution is t = 2 + 23S ≈ 5.46s. (2) according to the kinematic formula, x = V 0 (T + 2) = 4 × 7.46m = 29.84m (3) when the speed of the two is equal, the distance is the largest. Then t ′ = V 0A = 2S, the displacement of the bus: x 1 = V 0 (t ′ + 2) = 16m, the displacement of the motorcycle: x 2 = 12At ′ 2 = 4m, the maximum distance: △ x = x 1-x 2 = 12 m. A: (1) after the motorcycle starts, it overtakes the car in 5.46 seconds. (2) when the motorcycle overtakes the car, it is 29.84 meters away from the starting point. (3) before the motorcycle overtakes the car, the maximum distance between the two is 12 meters