Question

solution for this question​

Answer 1

Here, we are considering the acceleration to be constant.

The formula for distance travelled by nth second will be:

Sₙ =  u + $\frac{a}{2}$ (2n-1)

now we substitute all the values

u= 0, since the body was at rest

a= 5

n= 4

= 0 + $\frac{5}{2}$ (2(4)  - 1) = 2.5 x 7 = 17.5 m.

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Answer 2

Explanation:

S=ut+1/2at²

V=u+at

when t=1s

5=0+a

a=5m/s²

when t= 4s

5=0+4a

a=5/4=1.25m/s²

S1=0×1+1/2×5×1²

S2=0×4+1/2×1.25×4²

S1=2.5m

S2=10m

distance at the glance of 4th second:>

S2-S1=10-2.5=7.5m