Question

A body starts moving from rest and moves with a uniform acceleration of 5 cm/s?. In how much time the velocity of body will become 50 cm/s and how much will it travel in this line?​

Answer 1

Given :-

• The acceleration of body = 5cm/s
• The initial velocity of body = 0m/s
• The final velocity of body = 50cm/s

To Find :-

• The time taken by body (t) = ?
• Distance travelled by body = ?

Solution :-

• To calculate time and distance at first we have to convert acceleration into m/s² and final velocity into m/s then calculate it's time taken and distance travelled by the car by applying formula.

⇢Final velocity (v) = 50/100 = 0.5m/s

⇢Acceleration (a) = 5/100 = 0.05m/s²

Calculation for time :-

• As per 1st equation of motion :-]

⇢ v = u + a × t

⇢ 0.5 = 0 + 0.05 × t

⇢ 0.5 = 0.05t

⇢ t = 10s

Calculation for distance :-

• As per 2nd equation of motion :-]

⇢ S = u × t + 1/2 × a × t²

⇢ S = 0 × 10 + 1/2 × 0.05 × (10)²

⇢ S = 0 + 1/2 × 0.05 × 100

⇢ S = 1/2 × 0.05 × 100

⇢ S = 0.05 × 50

⇢ S = 2.5m

Hence, distance travelled by body = 2.5m

Hence, time taken by body = 10 seconds

Answer 2

$\large\text{Given:}$

$\rightarrowtail \bold{Acceleration (a) = 5 cm/s}$

$\rightarrowtail \bold{Final\;velocity(v)=50cm/s}$

$\rightarrowtail \bold{Initial\;velocity(u)=0cm/s}$

• ∵ Car is moving from rest.

$\large\text{To find:}$

$\rightarrowtail\bold{Time\;taken(t)}$

$\rightarrowtail \bold{Distance\;covered(S)}$

$\large\text{Solution:}$

• Using the first equation of motion.

$\mapsto\bold{v=u+at}$

• Where we have.

$\twoheadrightarrow\bold{v=\dfrac{50}{100}m/s }$

$\twoheadrightarrow\bold{a=\dfrac{5}{100}m/s^{2} }$

$\twoheadrightarrow\bold{u=0m/s}$

• Substituting we get.

$\nrightarrow \bold{0.5 = 0 + 0.05(t)}$

$\nrightarrow \bold{0.5=0.05t}$

$\nrightarrow\bold{t=\dfrac{0.5}{0.05} }$

∴ Time taken (t) is 10 seconds.

• Using the second equation of motion.

$\mapsto \bold{S=ut+\dfrac{1}{2}at^{2} }$

• Where we have.

$\twoheadrightarrow \bold{u=0m/s}$

$\twoheadrightarrow \bold{a=0.05m/s^{2}}$

$\twoheadrightarrow\bold{t=10sec}$

• Substituting we get.

$\nrightarrow \bold{S=0(10)+\dfrac{1}{2}(0.05)(10)^{2} }$

$\nrightarrow \bold{S=0+0.025(100)}$

∴ The distance covered is 2.5 meters.