The car starts to drive on the straight road from the standstill, the acceleration of the car changes with time within 60 seconds, and the distance of the car within 60 seconds is calculated

The car starts to drive on the straight road from the standstill, the acceleration of the car changes with time within 60 seconds, and the distance of the car within 60 seconds is calculated

(1) draw the V-T diagram of the vehicle in 0 ~ 60s;
(2) find out the distance of the car in 60 seconds
(1) the speed image is shown in the figure. (2) 900m
2m／s2 X 10s=20m／s
（20m／s2）2／2 X 2m／2=100m
20m／s X 30s=600m
20m／s X 20s-1／2 X 1m／s2 X （20s）2=200m
S total = 100m + 600m + 200m = 900m

The train moves along the straight track at the speed of V = 54km / h. for some reason, it needs to stop at a small station for 1 min, the increment A1 = 0.3m/s and A2 = 0.5m/s and 178; when the train stops at the station, and the increment A2 = 0.5m/s and 178; after the train starts, until the original speed is restored. The delay time Δ t of the train due to the suspension is calculated

From acceleration to deceleration, or from acceleration to acceleration, the speed changes from v = 54km / h (15m / s) to 0. In the process of change, the speed is equivalent to half of the original speed, that is, the driving speed of 7.5m/s (which is equivalent to doubling the time), so we only need to calculate the time required for acceleration and deceleration

The initial speed of the train at the station is 54km / h, and it will decelerate to 36km / h after 2S,
1) Acceleration of the train as it enters the station
2) The displacement of the train passing through the station after 10s

（1）54Km/h=54*5/18=15m/s 36Km/h=36*5/18=10/m/s
v1-v0=at 10-15=a*2 a=-2.5m/s^2
(2) When the speed is reduced to 0, the time is t = V0 / a = 15 / 2.5 = 6S
So the car is stationary at 6S, and the displacement at 10s is:
s=v0t/2=15*6/2=45m.