Question

A ship of mass 4×10^7 kg is pulled by a force of 6×10^5 newton. Find acceleration. If initially the ship was at rest, find its velocity after 10 minutes.

Answer 1

Mass= 4*10^7 kg.
Force = 6*10^5 N.

Force = mass*acceleration

$acceleration = \frac{mass}{force} \\ acceleration = \frac{ {4 \times 10}^{7} }{ {6 \times 10}^{5} } \\ = \frac{ {2 \times 10}^{7 - 5} }{3} \\ = \frac{ {2 \times 10}^{2} }{3} \\ = \frac{200}{3}m {s}^{ - 2}$
Initial velocity=0 m/s^2.
Time = 10 minutes = 10*60 = 600 seconds.

From 1st equation of motion:

●v=u+at.

When u=0:

●v=at.

$v = \frac{200}{3} \times 600 \\ = \frac{200 \times 600}{3} \\ = \frac{120000}{3} \\ = 4000m {s}^{ - 1}$

Answer 2

Hola ⚘⚘

Given:

• A ship of mass 4×10^7 kg is pulled by a force of 6×10^5 newton.

Tofind:

• Find acceleration. If initially the ship was at rest, find its velocity after 10 minutes.

Calculation:

• F = m*a

>> a = m/f

>> a = 4*10⁷/6*10⁵

>> a = 2*10^{7-5} / 3

>> a = 2*10²/3

>> a = 200/3 ms^{-2}

• ( Initial velocity = 0)

• ( Time = 10*60 = 600s)

• v = u+at

>> v = 0 + 200/3 * 600

>> v = 200*600/3

>> v = 120000/3

>> v = 4000 ms-1

Extra Information:

equations of motion

• s = ut+½at²
• s = vt- ½at²
• v = u+ at
• u = v-at
• v²-u² = 2as
• v²-u²/2s = a
• v²-u²/2a = s

where,

• s = distance covered
• v = final velocity
• u = initial velocity
• t = time taken
• a = accelartion