Question

A particle starts from rest with a constant acceleration. At a time t seconds, the speed is found to be 100m/s and 1second later the speed becomes 150m/s. The distance travelled during the t+1 th second is 25×n meters. Then n is​

Answer 1

Answer:

initial velocity(u)=0

acceleration(a)= a

time(t)= T

final velocity(v)= 100 m/s

v= u+at

100 = 0+aT

aT = 100 --------eq(1)

when t = T+1

v=150 m/s

v= u+at

150= 0+a(T+1) = aT+a

150= 100+a      [from eq(1)]

a=50

50×T =100         [from eq(1)]

T= 2s

distance traveled in nth second=

Sn= u + 1/2a [2n-1]

25n = 0+ 50/2 [2(T+1)-1] = 25 × [2(2+1)-1]

25n = 25 × (2×3) -1 = 25×5

25n = 125

n = 125/25= 5

value of n is 5m

hope this helps

Answer 2

Answer:

5

Step-by-step explanation:

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