Question

A particle starting from rest moves with constant acceleration. If it takes $ 5.0\sec $ to reach the speed $ 18km/h $ : find (a) average velocity during this period, and (b) the distance travelled by the particle during this period.

Answer 1

Hint : To solve this question, first we will find the constant acceleration of the particle by the help of First law of motion and then with the help of acceleration value and the given time taken, we can find the average velocity of the particle. With the distance formula, we can find the distance travelled by the particle.

Complete Step By Step Answer:
According to the question, when a particle starts from the rest with constant acceleration then it’s Initial velocity is $ 0m/s $ .
Initial Velocity, $ u = 0m/s $
And, as we can see in question, the maximum speed achieved by the particle is $ 18km/h $ :
Final Velocity, $ v = 18km/h = 18 \times \dfrac{5}{{18}}m/s = 5m/s $
The time taken to achieve the final velocity, $ t = 5\sec $ .
(a). We have to find the average velocity during the period of initial velocity to achieve the final velocity:
From first law of motion:
 $ v = u + at $
here, $ v $ is the final velocity.
 $ u $ is the initial velocity.
 $ a $ is the acceleration of the particle, and
 $ t $ is time taken by the particle.
 $ \because v = u + at \\
   \Rightarrow 5 = 0 + a \times 5 \\
   \Rightarrow 5 = 5a \\
   \Rightarrow a = 1m/{s^2} \\ $
So, the acceleration is $ 1m/{s^2} $ .
Now, to find the average velocity, we have to find the total displacement:
So, the distance is:
 $ \therefore s = ut + \dfrac{1}{2}a{t^2} \\
   \Rightarrow s = (0 \times 5) + \dfrac{1}{2}.1 \times {5^2} \\
   \Rightarrow s = 12.5m \\ $
Now, the average velocity during the period:
 $ \therefore Avg.\,Velocity = \dfrac{{Total\,displacement}}{{Total\,time\,taken}} \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{12.5}}{5} = 2.5m/s \\ $
(b). the distance travelled by the particle during this period:
 $ Dis\tan ce\,travelled = Average\,Velocity \times time\,taken \\
   \Rightarrow s = 2.5 \times 5 = 12.5m \\ $

Note :
There are two conditions on which the first law of motion is dependent: Objects at rest: When an object is at rest velocity $ v = {\text{ }}0 $ and acceleration $ a = 0 $ are zero. Therefore, the object continues to be at rest.