Question

3. A train starting from rest attains a velocity of 20m/sec. in 2minutes. Assuming that the acceleration is uniform, find (i) the acceleration. (ii) distance travelled by the train, while it attained this velocity.

Answer 1

Answer:

The acceleration of the train is $0.167m/s^2$ and distance traveled by the train is 1202 m.

Explanation:

Given the initial velocity of the train, u = 0.

The final velocity of the train, v = 20 m/s.

Time taken to attain the final velocity, t = 2 min = 120 s.

(i) To calculate the acceleration, use first equation of motion as  

v = u + at

Substitute the given values, we get

20 m/s = 0 +a x 120 s

$a=\frac{20m/s}{120s}$

$a=0.167m/s^2$

(ii) To calculate the distance traveled, use second equation of motion as,

$s=ut+\frac{1}{2}at^2$

substitute the values, we get

$s=0\times120s+0.5\times0.167m/s^2(120s)^2$

s = 1202.4 m

Thus, the acceleration of the train is $0.167m/s^2$ and distance traveled by the train is 1202 m.

#Learn More: Equations of motion.

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Answer 2

Explanation:

The acceleration of the train is 0.167m/s^20.167m/s

2

and distance traveled by the train is 1202 m.

Explanation:

Given the initial velocity of the train, u = 0.

The final velocity of the train, v = 20 m/s.

Time taken to attain the final velocity, t = 2 min = 120 s.

(i) To calculate the acceleration, use first equation of motion as

v = u + at

Substitute the given values, we get

20 m/s = 0 +a x 120 s

a=\frac{20m/s}{120s}a=

120s

20m/s

a=0.167m/s^2a=0.167m/s

2

(ii) To calculate the distance traveled, use second equation of motion as,

s=ut+\frac{1}{2}at^2s=ut+

2

1

at

2

substitute the values, we get

s=0\times120s+0.5\times0.167m/s^2(120s)^2s=0×120s+0.5×0.167m/s

2

(120s)

2

s = 1202.4 m

Thus, the acceleration of the train is 0.167m/s^20.167m/s

2

and distance traveled by the train is 1202 m.