Question

A body with an initial velocity of 4m/s accelerates uniformly at the rate of 1.2 m/s² over a distance of 20 m. Calculate: (i) the final velocity of the body.(ii) the time in which the body covers that distance. (iii) the average velocity of the body during the journey.

Answer 1

Given :

• Initial velocity (u) = 4 m/s
• Acceleration (a) = 1.2 m/s²
• Distance (s) = 20 m

To find :

• Final velocity (v)
• Time taken (t)
• Average velocity

According to the question,

i)

$\leadsto { \boxed{\sf \red{ {v}^{2} = {u}^{2} + 2as }}}$

$\leadsto \sf{ {v}^{2} = 4 {}^{2} + 2 \times 1.2 \times 20 }$

$\leadsto \sf{ {v}^{2} = 16 + 2 \times \frac{12}{10} \times 20}$

$\leadsto \sf{ {v}^{2} = 16 + 48}$

$\leadsto \sf{ {v}^{2} = 64}$

$\leadsto \sf{v = 8 \: m {s}^{ - 1} } \green\bigstar$

So,the final velocity is 8 m/s...

ii)

$\leadsto {\boxed{\sf \red{v = u + at}}}$

$\leadsto \sf{8 = 4 + 1.2 \times t }$

$\leadsto \sf{8 - 4 = 1.2t}$

$\leadsto \sf{4 = \dfrac{12}{10} t}$

$\leadsto \sf{ \dfrac{4 \times 10}{12} = t }$

$\leadsto \sf{3.33 = t} \: \green \bigstar$

So,the time is 3.33 seconds..

iii)

$\leadsto{\boxed {\sf \red{Average \: velocity = \frac{Displacement}{Time} }}}$

$\leadsto \sf{ \frac{20}{3.33} }$

$\leadsto \sf{6.006 \: {ms}^{ - 1} } \green\bigstar$

So,the average velocity is 6.006 m/s..