Alight-rail commuter train accelerates at a rate of 1.35 m/s2. how long does it take to reach its top speed of 80.0 km/h, starting from rest? (b) the same train ordinarily decelerates at a rate of 1.65 m/s2. how long does it take to come to a stop from its top speed? (c) in emergencies the train can decelerate more rapidly, coming to rest from 80.0 km/h in 8.30 s. what is its emergency deceleration in m/s2?

a) 16.46 seconds.

b) 13.46 seconds

c) 2.67 m/s²

Explanation:

Acceleration = a = 1.35 m/s²

Final velocity = v = 80 km/h =

Initial velocity = u = 0

Equation of motion

Time taken to accelerate to top speed is 16.46 seconds.

Acceleration = a = -1.65 m/s²

Initial velocity = u = 80 km/h=

Final velocity = v = 0

Time taken to stop the train from top speed is 13.46 seconds

Initial velocity = u = 80 km/h=

Time taken = t = 8.3 s

Final velocity = v = 0

Emergency deceleration is 2.67 m/s²

dec 23, 2016 - a load of bricks weighing 680 n a distance of 10 m across a parking lot. if if the worker pushes the cart with a constant force of 209 n, what work= force x distance = 220 n x 10 m = 2200 n.m = 2200 j i don't know where the person got it from, but the initial part, work= force x distance me.

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