Question

A body travelling at uniform retardation 2.5m/s2 has its velocity reduced to
half the initial value in travelling a distance of 540m. Calculate the initial
velocity and the time interval to travel distance of 540.

Answer 1

Given:-

• Distance travelled = 540

• Retardation = 2.5m/s²

• Initial Velocity = u

• v = u / 2

To Find:-

• Initial Velocity and time taken to cover 540m.

Formulae used:-

• v² - u² = 2as

• v = u + at

Where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• s = Distance
• t = Time.

Now,

→ v² - u² = 2as

→ (u/2)² - (u)² = 2 × -2.5 × 540

→ u²/4 - u² = 2700

→ -3u²/4 = 2700

→ -3u² = 2700 × 4

→ -3u² = -10800

→ -u² = 10800/3

→ -u² = -3600

→ √-u² = √-3600

→ u = 60m/s

Hence, The Initial Velocity of body is 60m/s.

Now, To find Time,

→ v = u/2

→ v = -60/2

→ v = 30m/s

Now,

→ v = u + at

→ 30 = 60 + -2.5 × t

→ 30 - 60 = -2.5 × t

→ -30 = -2.5t

→ t = 30/2.5

→ t = 12s.

Therefore, The Time taken to cover 540m is 12s.

Answer 2

Answer:

Correct Question:

A body travelling at uniform retardation 2.5m/s^2 has its velocity reduced to half the initial value in travelling a distance of 540m. Calculate the initial

velocity and the time taken to travel distance of 540 m.

Solution:

We need to find initial velocity and time taken to cover the distance of 540 m.

We know that,

$v {}^{2} - u {}^{2} = 2as \\ \\ = > ( \frac{u}{t} ) {}^{2} - u {}^{2} = 2as \\ \\ = > \frac{u {}^{2} }{4} - u {}^{2} = 2 \times - 2.5 \times 540 \\ \\ = > \frac{u {}^{2} }{4} - u {}^{2} = - 2700 \\ \\ = > \frac{u {}^{2} - 4u {}^{2} }{4} = - 2700 \\ \\ = > \frac{ - 3u {}^{2} }{4} = - 2700 \\ \\ = > - 3u {}^{2} = - 2700 \times 4 \\ \\ = > - 3u {}^{2} = - 10800 \\ \\ = > - u {}^{2} = - \frac{10800}{3} \\ \\ = > - u {}^{2} = - 3600 \\ \\ = > \sqrt{ - u {}^{2} } = \sqrt{ - 3600} \\ \\ = > u = - 60 \: \frac{m}{s}$

Initial velocity is - 60 m/s.

Now we need to find the time taken to travel the distance of 540 m.

v = u/t

= -60/2

= - 30m/s

v = u + at

$= > - 30 = - 60 + ( - 2.5) \times t \\ \\ = > - 30 + 60 = - 2.5t \\ \\ = > 30 = - 2.5t \\ \\ = > 2.5t = 30 \\ \\ = > t = \frac{30}{2.5} \\ \\ = 12s$

Therefore, initial velocity of the body is 60m/s and time taken to cover the distance of 540m is 12s.