Question

. A particle starts with an initial velocity 3.0 m/s along the positive x-direction and is accelerated uniformly at rate of 0.60m/s2. It will take __________________ time and will cover_____________ distance to reach the velocity 9.0 m/s2.​

Answer 1

*★ Given :

initial velocity (u) = 3 m/s

acceleration (a) = 0.6 m/s²

final velocity (v) = 9 m/s²

★ To find :

Time taken and the distance covered...

★ Solution :

We know that,

✏️ v = u + at

✏️ v² = u² + 2as

★ To find time taken :

v = u + at

by putting values

9 = 3 + 0.6 t

9 - 3 = 0.6 t

6 = 0.6 t

t = 6/0.6

→ t = 10 s

★ To find distance covered :

v² = u² + 2as

by putting values

(9)² = (3)² + 2 (0.6) s

81 = 9 + 1.2 s

81 - 9 = 1.2 s

72 = 1.2 s

s = 72/1.2

s = 720/12

→ s = 60 m

Answer 2

Solution :

$\underline{\bf{Given\::}}}}$

• Initial velocity, (u) = 3.0 m/sec.
• Final velocity, (v) = 9.0 m/sec.
• Acceleration, (a) = 0.60 m/sec²

$\underline{\bf{To\:find\::}}}}$

The time cover & distance cover of the particles .

$\underline{\bf{Explanation\::}}}}$

We know that formula of the first equation of motion using for get time :

$\boxed{\bf{v=u+at}}}$

$\longrightarrow\sf{9=3+0.6\times t}\\\\\longrightarrow\sf{9=3+0.6t}\\\\\longrightarrow\sf{0.6t=9-3}\\\\\longrightarrow\sf{0.6t=6}\\\\\longrightarrow\sf{t=\dfrac{6\times 10}{0.6\times 10}} \\\\\longrightarrow\sf{t=\cancel{60/6}}\\\\\longrightarrow\bf{t=10\:seconds}$

Now, using third equation of motion for distance cover :

$\boxed{\bf{2as=v^{2}-u^{2}}}}$

$\longrightarrow\sf{2\times 0.6\times s=(9)^{2}-(3)^{2}}\\\\\longrightarrow\sf{1.2s=81-9}\\\\\longrightarrow\sf{1.2s=72}\\\\\longrightarrow\sf{s=\dfrac{72\times 10}{1.2\times 10} }\\\\\longrightarrow\sf{s=\cancel{720/12}}\\\\\longrightarrow\bf{s=60\:m}$

Thus;

The time taken 10 sec. & distance covered will 60 m .