Question

Example 3.6
A particle having initial velocity u moves with a constant
acceleration a for a time t. (a) Find the displacement of
the particle in the last 1 second. (b) Evaluate it for
u = 5 m/s, a = 2 m/s ? and t = 10 s.​

Answer 1

In first part of the question we have to find the displacement of the particle in the last 1 second.

a) Using the Third Equation Of Motion

In last one second, time (t) = (t - 1) sec,

s' = u(t - 1) + 1/2 a(t - 1)² .......(1st equation)

Displacement of the particle at the end of 't' second

s" = ut + 1/2 at² ................(2nd equation)

S = s" - s'

Substitute value of (1st equation) and (2nd equation)

S = ut + 1/2 at² - [u(t - 1) + 1/2 a(t - 1)²]

S = ut + 1/2 at² - [ ut - u + 1/2 a(t² + 1 - 2t) ]

S = ut + at²/2 - ut + u - at²/2 - a/2 + at

S = u - a/2 + at

S = u - a (1/2 - t)

S = u + a (-1/2 + t)

S = u + a/2 (2t - 1)

b) In second part of the question we have given values, we just have to put them.

u = 5 m/s, a = 2 m/s² and t = 10 s.

S = 5 + 2/2 (2*10 - 1)

S = 5 + 1(20 - 1)

S = 5 + 19

S = 24

Therefore, distance covered by the particle is 24 m.

Answer 2

Given:

• Initial velocity of the particle : u m/s
• It has a constant acceleration of : a m/s²
• For time : t

To find:

• a) Displacement in the last second
• b) Evaluation when
• u = 5m/s , a = 2m/s & t = 10s

Solution:

Here , we need to find Displacement in last second so , time in last second

→ Total time - 1s

→ t = t - 1

a) sol:

Now, using the kinematic equation

s = ut + 1/2 at²

Substituting t = t - 1

→ s' = u(t - 1) + 1/2 a(t - 1)²

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→ s" = ut + 1/2 at²

Now,

S = S" - S'

Substituting the values we have

→ S = ut + 1/2 at² - [ u(t - 1) + 1/2 a(t - 1)² ]

→ S = ut + 1/2 at² - { ut - u + 1/2 a[(t² + 1² - 2t)] }

→ S = ut + 1/2 at² - ut + u - 1/2 (at² + a - 2at)

→ S = 1/2at² + u - 1/2at² + a/2 - 2at/2

→ S = u + a/2 - at

→ S = u + a/2(2t - 1)

Hence ,

Displacement at last second = u + a/2(2t - 1).

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b) sol:

Now, just substituting the given values we got in the above question

→ S = 5 + 2/2[2(10) - 1]

→ S = 5 + 20 - 1

→ S = 24

Hence, distance travelled = 24m