Question

A particle is projected vertically upward with the velocity of 40 metre per second find the displacement and distance travelled by the particle in 1) 2 sec.2) 4sec. 3) 6sec​

Answer 1

Answer:

u = 40 m/s

Hmax = u² / 2g

          = 1600 / 20

          = 80 m

time taken to reach maximum height ,

t = u / g

 = 40 / 10

 = 4 sec

• at t = 1 ,

S = 40 + 1/2 × 10 ×1

  = 45 m

distance = displacement = 45 m

• at t = 2 sec

S = 40 + 1/2 × 10 × 4

  = 60 m

distance = displacement = 60 m

• at t = 4s

S = 40 + 1/2 × 10 × 16

  = 80 m

Distance = displacement = 80 m

• at t = 6 sec ,

Note : the body would be falling freely . Since it takes 4 sec to reach maximum height , We have to find S with t = 2 and tthen add it with t = 4

For distance ,

u = 0

t = 2 sec

a = g = 10 m/s²

S = 1/2 gt²

  = 1/2 × 10 × 4

  = 20 m

The displacement is 20 m from the top and 60 m from the bottom

for distance ,

S = 40 + 20

   = 60 m

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Answer 2

Answer:

Explanation:-

Distance covered is equal to displacement if the object moves in a straight line and there is no change in direction of motion. If direction of motion changes, distance should be calculated separately for different parts of the path. Here, due to upward motion, u is positive and due to downward motion, a is negative. Velocity becomes zero at maximum height Time taken to reach maximum height (t°) = u/g

So,t=u/g= 40÷10=4=t°

Now,distance and displacement are:-

a) from Newton's second law of motion

S= 60 m

b) similarly, at t=t°, distance and displacement would be equal,so

from Newton's second law of motion

S= 80m

C)

As v=u+at, t = v−ua = 0–40−10=4s

The distance covered in this time, sup=u∗t+12a∗t2

sup=40∗4−10∗8=80m (Same reason as above for the negative sign)

Now, let us analyze the downward journey.

Time, t=6–4=2s

Sdown=u∗t+12a∗t2=0+20=20m The distance covered in this time,

So, total distance covered =sup+sdown=80+20=100m

On the other hand, the displacement is the shortest distance between the initial and final point i.e., the shortest distance between the ground and the position after 6s of travel. Hence, displacement =80–20=60m.