Question

44. A train starts from station with an acceleration 1 m/s2.
A boy who is 48 m behind the train with a constant
velocity 10 m/s, the minimum time after which the boy
will catch the train is
(A) 4.8 sec
(B) 8 sec
(C) 10 sec
(D) 12 sec​

Answer 1

Answer:

(a) 4.8 sec the boy will catch the train

Explanation:

This will help u

Answer 2

✯ The minimum time = 8 sec ✯

Explanation:

Given:

• Acceleration of train = 1 m/s²
• Velocity of boy = 10 m/s
• Distance between the train and boy = 48 m.

To find:

• The minimum time after which the boy will catch the train.

Solution:

Acceleration of the boy = 0 m/s²

Then,

Acceleration of boy w.r.t train = Acceleration of boy - Acceleration of train

= (0 - 1) m/s²

= -1 m/s²

• a = -1 m/s²

We know that,

✯ s = ut + ½ at² ✯

Here,

• s = 48 m
• u = 10 m/s
• t = ?
• a = -1 m/s²

[ Put values ]

$\implies$ 48 = 10×t + ½ ×(-1)× t²

$\implies$ 48 = 10t - t²/2

$\implies$ 48 = (20t-t²)/2

$\implies$ 20t - t² = 96

$\implies$ -t² + 20t -96 = 0

$\implies$ t² - 20t + 96 = 0

$\implies$ t² - 12t - 8t +96 = 0

$\implies$ t (t-12) - 8(t-12) = 0

$\implies$ (t-12) (t-8) = 0

Either,

t - 12 = 0

→ t = 12

Or,

t - 8 = 0

→ t = 8

Therefore, the minimum time after which the boy will catch the train is 8 sec.

† Option (B) 8 sec is correct.

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