Question

10. A particle starts from rest, moves with constant
acceleration of 10 m/s2. If it travels 35 m in the nth
second of its motion, then the total distance
covered by the particle in n second is
(1) 20 m
(2) 40 m
(3) 60 m
(4) 80 m
11. A particle moves in a straight line such that its
position is given by x = 5(t-2) +6t2, where x is
in meter and t is in second. The initial velocity of
particle is
(1) Zero
(2) 2 m/s
(3) 5 m/s
(4) 12 m/s

e moving with initial velocity to​

Answer 1

For question 10 as we know distance travelled in nth second is

u+a/2(2n-1)

as u =0

and a=10m/s²

we have

5(2n-1)=35

2n-1=7

2n=8

n=4seconds

now distance trave in n second is

s=ut+1/2at²

as u=o

s=5×4²

s=80m

so for question 10 your answer is d 80m

For question 11make sure your question is correct.what does "e moving with initial velocity to" mean please write it correctly

Answer 2

answer

1...80m is the correct answer

explanation

we know that

Distance travel in nth second is

•••u+a/2(2n-1)

as u =0

and a=10m/s²

we have

5(2n-1)=35

2n-1=7

2n=8

n=4 sec

now,

distance travelled in n second is

s=ut+1/2at²

as u=o

s=5×4^2

s=80m

answer

2.).....5m/s is the correct answer......

explanation

Considering the equation is x = 2 +5t +6t²

x= 6t² + 5t + 2

We know that,

v=dx / dt

So,

v= d(6t² +5t +2) /dt

v= 12t + 5

For, initial velocity substitute t=0;

v= 12(0) +5

v= 5m/sec

So, the initial velocity was 5m/sec

I hope it help u dear