# Train Question....!

• Sep 24th 2006, 02:17 AM
classicstrings
Train Question....!
Q: A train runs on a straight track between two stops, A and B, a distance of four kilometers. It accelerates from rest at A with acceleration of 0.3 m/s2 until it reaches the maximum speed of 24 m/s (meters per second). It maintains this speed for a time and then decelerates to rest at B at a rate of 0.5 m/s2. Find the time taken for the journey.

Q: A bus starts from rest and moves with a constant acceleration until it reaches a speed of 18 m/s. It continues at this speed for a time, after which it is brought to rest with constant deceleration. The total time taken is 24 seconds and the distance travelled by the bus is 288 metres. If the time for the deceleration is half that for the acceleration, find:

a) the time the bus takes to decelerate from its top speed to rest.
b) the distance the bus travels at its top speed of 18 m/s.

Not sure how to tackle these questions, thank you!
• Sep 24th 2006, 02:40 AM
Glaysher
Quote:

Originally Posted by classicstrings
Q: A train runs on a straight track between two stops, A and B, a distance of four kilometers. It accelerates from rest at A with acceleration of 0.3 m/s2 until it reaches the maximum speed of 24 m/s (meters per second). It maintains this speed for a time and then decelerates to rest at B at a rate of 0.5 m/s2. Find the time taken for the journey.

Sketching a speed-time graph will help. The gradient is the acceleration and the area underneath is the distance travelled

24/0.3 = 80 seconds to accelerate

(80 * 24)/2 = 960m travelled

24/0.5 = 48 seconds to deccelerate

(48 * 24)/2 = 576m

Total travelled whilst accelerating/deccelerating = 1536m

Total left to travel = 4000 - 1536 = 2464m

At 24m/s time taken = 2464/24 = 102 2/3

Time taken = 80 + 48 + 102 2/3 = 230 2/3 seconds
• Sep 24th 2006, 02:53 AM
Glaysher
Quote:

Originally Posted by classicstrings

Q: A bus starts from rest and moves with a constant acceleration until it reaches a speed of 18 m/s. It continues at this speed for a time, after which it is brought to rest with constant deceleration. The total time taken is 24 seconds and the distance travelled by the bus is 288 metres. If the time for the deceleration is half that for the acceleration, find:

a) the time the bus takes to decelerate from its top speed to rest.
b) the distance the bus travels at its top speed of 18 m/s.

Not sure how to tackle these questions, thank you!

Again draw a speed time graph

a) 288 = 18t/2 + (18 * 0.5t)/2 + 18(24 - 1.5t)

288 = 9t + 4.5t + 432 - 27t

288 = -13.5t + 432

13.5t = 144

t = 10 2/3

Half t for answer 5 1/3

b) 18(24 - 1.5 * 10 2/3) =