Train speed
The train speed is decreased from 72 km/h to 36 km/h in 50 seconds. If the train movement is continuously slowing, find the acceleration and the distance it travels.
Final Answer:
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Dr. Math
Solution:
We are given:
- Initial speed, u = 72 km/h ,
- Final speed, v = 36 km/h ,
- Time, t = 50 seconds .
Step 1: Convert speeds from km/h to m/s.
To convert from km/h to m/s, multiply by 10003600 = 518 :
Step 2: Calculate acceleration.
Acceleration ( a ) is the rate of change of velocity. Since the train is slowing down, the acceleration will be negative (deceleration). The formula for acceleration is:
Substitute the values:
Step 3: Calculate the distance traveled.
The distance ( s ) traveled under constant acceleration can be calculated using the formula:
Substitute the values:
Final Answers:
- The acceleration is:
- The distance traveled is:
We are given:
- Initial speed, u = 72 km/h ,
- Final speed, v = 36 km/h ,
- Time, t = 50 seconds .
Step 1: Convert speeds from km/h to m/s.
To convert from km/h to m/s, multiply by 10003600 = 518 :
u = 72 × 518 = 20 m/s
v = 36 × 518 = 10 m/s
Step 2: Calculate acceleration.
Acceleration ( a ) is the rate of change of velocity. Since the train is slowing down, the acceleration will be negative (deceleration). The formula for acceleration is:
a = v - ut
Substitute the values:
a = 10 - 2050 = -1050 = -0.2 m/s2
Step 3: Calculate the distance traveled.
The distance ( s ) traveled under constant acceleration can be calculated using the formula:
s = ut + 12 a t2
Substitute the values:
s = 20 × 50 + 12 × (-0.2) × 502
s = 1000 + 12 × (-0.2) × 2500
s = 1000 - 250
s = 750 m
Final Answers:
- The acceleration is:
-0.2 m/s2
- The distance traveled is:
750 m
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