Question

A point initially at rest moves along x-axis. Its

acceleration varies with time as

2

a (6t 5) m/s .

If it starts from origin, the distance covered in 2 s

is:

(a) 20 m (b) 18 m

(c) 16 m (d) 25 m​

Answer 1

Given :-

u = 0

a = ( 6t + 5) m/s^2

t = 2 sec

To find :-

the distance covered in 2 sec

Solution :-

∴ $= a = \frac{dv}{dt}$

= 6t + 5

∴ dv = ( 6t + 5 ) dt

taking integration ,

we have

= $\int\limits^v_0 \, dv = \int\limits^t_0 {( 6t + 5 )} \, dx$

v = $3t^{2} + 5t +c$

where c , is constant of intrigation

when ,

t = 0 , v = 0 , so  c = 0

   ∴ ds/dt = 3t^2 + 5t

or ds = ( 3t^2 + 5t)dt

so,

Intrigation it withine the conditions of s changes from 0 to s

we have ,

$\int\limits^s_0 \, dx = \int\limits^2_0 { 3t^2 + 5t} \, dx$

∴$s = t^{3} + \frac{5}{2} t^{2}$

= 8 +10

= 18 m

thank you

Answer 2

Answer:

Step-by-step explanation:

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