Posted by **PLZ HELP ME NOW...** on Wednesday, May 30, 2012 at 6:03am.

a train starts from rest and accelerates uniformly to achieve a velocity of 20m/sec in 10 sec.then train maintains this speed for next 200 sec.the breaks are then applied and train comes to rest in next 50 sec.calculate

a)acceleration in first 10 sec.

b)acceleration in last 50 sec.

c)the total distance traveled by the train in whole journey

d)average velocity of the trip.

- physics -
**drwls**, Wednesday, May 30, 2012 at 6:35am
a) (Velocity change)/Time = 2.0 m/s^2

b) (Velocity change)/Time = -0.4 m/s^2

c) X1 + X2 + X3, where X1 is the distance travelled accelerating; X2 is the distance travelled at constant speed and X3 is the distance travelled while decelerating.

= 10*10 + 20*200 + 10*50 = ___ m

Average speed was used for X1 and X3.

d) Divide previous answer by total elapsed time, 260 s.

## Answer This Question

## Related Questions

- physics - a train starts from rest and accelerates uniformly to achieve a ...
- physics - a train starts from rest and accelerates uniformly to achieve a ...
- physics - a body moves with a uniform velocity of 2m/sec for 5 sec then velocity...
- physics - a train starts from rest and accelerates uniformly at 100m/min^2 for ...
- physics - a train starts from rest and accelerates uniformly at 100m/min^2 for ...
- physics - a train starts from rest and accelerates uniformly at 100m/min^2 for ...
- physics - a cyclist start from rest and accelerate uniformly to achieve a ...
- Physics - the engineer of a passenger train traveling at 100 ft/sec sights a ...
- physics - a car traveled at uniform velocity of 20m/sec for 5 sec the brakes are...
- physics - a car starts from rest and attains a velocity of meter/sec,then ...

More Related Questions