Question

A body with an initial velocity of 18 km/h accelerates uniformly at the rate of 10 cm/s2 over a distance of 300 m, calculate: i) The acceleration in m/s2 ii) Its final velocity in m/s.​

Answer 1

$\\ \huge{\underline{\underline{\sf{\pmb{Question:}}}}}$

A body with an initial velocity of 18 km/h accelerates uniformly at the rate of 10 cm/s2 over a distance of 300 m, calculate:

• i) The acceleration in m/s^(2)
• ii) Its final velocity in m/s.

$\\ \huge{\underline{\underline{\sf{\pmb{Required \: Answer:}}}}}$

$\\ \large{\underline{\underline{\sf{\frak{\bold{Given:}}}}}}$

• Initial velocity = 18km/h
• Acceleration = 10cm/s^(2)
• Distance = 300m

$\\ \large{\underline{\underline{\sf{\frak{\bold{To \: Find:}}}}}}$

• Acceleration in m/s^(2)
• Final velocity in m/s

$\\ \large{\underline{\underline{\sf{\frak{\bold{Solution:}}}}}}$

i) The acceleration in m/s^(2)

We know that:-

• 1cm = 0.01m

We were given,

• Acceleration = 10cm/s^(2)

Therefore,

$\\ \sf{Acceleration = \dfrac{10cm}{ {s}^{2}} } = \dfrac{(10 \times 0.01)m}{ {s}^{2} } = \bold{\red{0.1m {s}^{ - 2}}}$

Hence, the acceleration is 0.1m/s^(2)

ii) Final velocity in m/s

We know that:-

• $\\ \large{\underline{\boxed{\rm{\bold{ {v}^{2} = {u}^{2} + 2as}}}}}$

Where,

• v = Final velocity
• u = initial velocity
• a = Acceleration
• s = Distance.

We have,

• Initial velocity u = 18km/h
• Acceleration a = 10cm/s^(2) = 0.1m/s^(2)
• Distance s = 300m

Substituting the values:-

$\\ \bold{ {v}^{2} } = \tt{(18) {}^{2} + 2 \times 0.1 \times 300}$

$\\ \tt{\implies{ {v}^{2} = 324 + 60}}$

$\\ \tt{\implies{ {v}^{2} = 384}}$

$\\ \tt{\implies{v = \sqrt{384}}}$

$\\ \tt{\therefore{v = \red{\bold{19.6m/s}}}}$

Hence, final velocity of the body is 19.6m/s