Question

A particle moves in
straight line with a constant acceleration.It changes its velocity from 10m/s to 20m/s while passing through
a distance of 135m in t seconds. The value of t is______​

Answer 1

Given :

A particle moves in  straight line with a constant acceleration . It changes its velocity from 10m/s to 20m/s while passing through  a distance of 135m in t seconds

To Find :

Value of t

[ As the acceleration is constant throughout the motion in a straight line , we need to apply equations of kinematics  $\orange{\bigstar}$ ]

Solution :

$\bigstar\ \; \sf \purple{v^2-u^2=2as}\\\\\bigstar\ \; \sf \red{v=u+at}$

where ,

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• s denotes displacement / distance
• t denotes time

Given ,

Initial velocity , u = 10 m/s

Final velocity , v = 20 m/s

Distance , s = 135 m

Note : In a straight line , distance and displacement are both same

$\bigstar\ \; \pink{\sf v^2-u^2=2as}\\\\\to \sf (20)^2-(10)^2=2a(135)\\\\\to \sf 400-100=270a\\\\\to \sf 300=270a\\\\\to \sf a=\dfrac{300}{270}\\\\\to \sf a=\dfrac{30}{27}\\\\\leadsto \sf a=\dfrac{10}{9}\ m/s^2\ \; \bigstar$

$\bigstar\ \; \sf \blue{v=u+at}\\\\\to \sf 20=10+\left(\dfrac{10}{9}\right)t\\\\\to \sf 20-10=\dfrac{10t}{9}\\\\\to \sf 10=\dfrac{10t}{9}\\\\\to \sf 90=10t\\\\\to \sf 10t=90\\\\\leadsto \sf \orange{t=9\ s}\ \; \bigstar$

Answer 2

Answer:

Time = 9 second

Explanation:

Given that,

• Initial velocity (u) = 10 m/s
• Final velocity (v) = 20 m/s
• Displacement (s) = 135 m

Applying 3rd equation of motion,

✒ v² - u² = 2as

[ Where, v = final velocity, u = initial velocity, a = acceleration, s = displacement ]

➡ (20)² - (10)² = 2 × a × 135

➡ 400 - 100 = 270a

➡ 300 = 270a

➡ a = 300/270

➡ a = 10/9

➡ a = 1.1 m/s²

Now, Applying 1st equation of motion,

✒ v = u + at

[ Where, v = final velocity, u = initial velocity, a = acceleration, t = time ]

➡ 20 = 10 + 1.1 × t

➡ 20 = 10 + 1.1t

➡ 20 - 10 = 1.1t

➡ 10 = 1.1t

➡ t = 10/1.1

➡ t = 9 sec