Question

A particle starts with velocity 10 ms-1 and
constant retardation 2 ms-2. The distance
travelled in 5th second is
1 m
25 m
24 m
2 m​

Answer 1

Answer:

hello

Explanation:

The distance travelled in n seconds is

Sn=un+(1/2)an^2……..(1)

The distance travelled in (n-1) seconds,

S(n-1)=u (n-1)+(1/2)a(n-1)^2=un-u+(1/2)an^2-an+a/2………(2)

The distance travelled in nth second is

Sn-S(n-1)=un+(1/2)an^2-un+u-(1/2)an^2+an-(1/2)a=u+an-a/2=u+a(n-1/2)…..(3)

Now, in the given problem, initial velocity, u=10m/s, acceleration, a=-2m/s^2. Negative sign denotes retardation. n=5.

Substituting these values in equation (3), we have distance travelled in 5th second,

S5-S4=10–2(5–1/2)=10–2(9/2)=10–9=1m

Hope Helpful.....

Answer 2

Answer : i think the answer is 25 m .