Question

$\large\mathcal{\red{QUESTION :-}}$


Find distance travelled by the uniformly accelerated object moving in one dimension in nth second​.




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Answer 1

Step-by-step explanation:

Using the equation of motion, the position ‘x’ of a particle with an acceleration ‘a’ at time ‘t’ is given by

x[t] = ut + (1/2)at^2

So, position during ‘n’th second is

x[n] = un + (1/2)an^2

Similarly, position during ‘(n-1)’th second is

x[n-1] = u(n-1) + (1/2)a(n-1)^2

Distance travelled during ‘n’th second = Position change from ‘(n-1)’th second to ‘n’th second = x[n] - x[n-1]

= u + (a/2)(2n-1) …On simplification.

Answer 2

velocity

Displacement formula for motion

S=ut+

2

1

at

2

displacement after n seconds

S

n

=ut

n

+

2

1

at

n

2

Displacement after (n-1) seconds

S

n−1

=ut

n−1

+

2

1

at

n−1

2

Displacement in n

th

second is

s

n

th

=s

n

−s

n−1

⟹S

n

th

=u(t

n−1

−t

n

)+

2

1

a(t

n−1

−t

n−1

)

2

because (t

n−1

−t

n

)=1

Dimension of the S

n

th

is m/s