Question

A stone is thrown in a vertically upward direction with a velocity of 25ms-1 .If the acceleration of the stone is 10ms-2, in downward direction ,what will be the height attained by the stone and how much time will it take to reach there, (or ) A train starting from rest attains a velocity of 90 km.h-1 in 10 mins assuming that the acceleration is uniform , find (i) the acceleration and the distance travelled by the train for attaining this velocity

Answer 1

Answer - I :

Initial velocity of ball = 25 m/s

Acc. due to gravity = 10m/s²

We have to find

• Max height attained by the ball
• Time taken by ball to reach there

For a body thrown vertically upward, g is taken negative$.$

(i) Maximum height :

➝ v² - u² = 2gH

At maximum height, v = 0

➝ 0² - (25)² = 2(-10)H

➝ -625 = -20H

➝ H = 31.25 m

(ii) Time taken :

➝ v = u + gt

➝ 0 = 25 + (-10)t

➝ t = 25/10

➝ t = 2.5 s

Answer - II :

Initial velocity = zero

Final velocity = 90km/h = 25m/s

Time of journey = 10min = 600s

We have to find

• Acceleration of train
• Distance travelled by the train

(i) Acceleration :

We know that acceleration is defined as the rate of change of velocity.

➝ a = (v - u) / t

➝ a = (25 - 0) / 600

➝ a = 0.04 m/s²

(ii) Distance :

Distance covered by the train can be calculated by using third equation of motion.

➝ v² - u² = 2as

➝ 25² - 0² = 2(0.04)s

➝ s = 625/0.08

➝ s = 7812.5 m = 7.8 km

Answer 2

Ques-1

Given:

• Initial velocity (u) = 25 m/s
• Final velocity (v) = 0 m/s
• Acceleration in downward direction (-a) = -10 m/s^2

To find:

• Time to reach a height (t) = ?
• Height the stone will attain (s) = ?

Solution:

We know Newton's First kinematic equation i.e, v = u + at

For time put the values im the 1st equation of motion.

==> 0 = 25 + (-10) × t

==> -25 = -10 × t

==> -25/-10 = t

==> 2.5 s = t

Hence, the time taken = 2.5 seconds.

Now, we know time we can find the distance or height attained by the stone by Newton's 2nd equation of motion i.e, s = ut + 1/2×a×t^2

(Putting values)...

==> s = 25 × 2.5 + 1/2 × -10 × (2.5)^2

==> s = 62.5 - 5 × 6.25

==> s = 62.5 - 31.25

==> s = 31.25 m

Hence, the height stone will attain = 31.25 meters.

Ques-2

Given:

• Initial velocity (u) = 0 m/s
• Final velocity (v) = 90 km/h = 90 × 1000/3600 = 25 m/s
• Time (t) = 10 mins = 10 × 60 = 600 seconds

To find:

• Acceleration (a) = ?
• Distance traveled by the train (s) = ?

Solution:

We know the formula for acceleration i.e, a = v - u/t

(put the given values for acceleration)...

==> a = 25 - 0/600

==> a = 25/600

==> a = 0.04 m/s^2

Hence, the acceleration = 0.04 m/s^2.

Now we found acceleration, distance traveled can be calculated by Newton's 2nd kinematic equation of motion i.e, v^2 - u^2 = 2as.

(put the given values for distance traveled)...

==> (25)^2 - (0)^2 = 2 × 0.04 × s

==> 625 = 0.08 × s

==> 625/0.08 = s

==> 7812.5 m = s

Hence, the distance traveled by the train = 7812.5 meters.

Hope it helped you dear...