Question

A particle having velocity u moves wuth 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 = 5m/s, a = 2m/s^2 and t = 10s.​

Answer 1

Explanation:

initial velocity=u

acceleration= a

time=t

a). S=ut+1/2at^2

* Displacement at the.end of time (t-1)

S1=u(t-1)+1/2a(t-1)^2

* Displacement at the end of time t

S2=ut+1/2at^2

* displacement in last second

S2-S1

u+1/2at^2-1/2a(t-1)^2

u+a/2 (2t-1)

b). S=ut+1/2at^2

S=u+a/2×(2t-1)

S=5+2/2×(2×10-1)

S=5+1×19

S=24 m

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Answer 2

Answer:

st = ut + 1/2 at^2

st - 1 = u(t - 1) + 1/2 a(t - 1)^2

a) st - st - 1 = ut + 1/2 at^2 - {u(t - 1) + 1/2a(t - 1)^2}

= ut + 1/2at^2 - {ut - u + 1/2a(t^2 + 1 - 2t)}

= ut + 1/2 at^2 - ut + u - 1/2at^2 - a/2 + at

= u - a/2 + at

= u + a[t - 1/2]

b) ut + 1/2at^2

= 5 + 1 × 19

= 24m.