Question

A point initially at rest moves along x-axis. Its acceleration varies with time as a = (6t + 5)m / s² . 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

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Answer 2

Answer:

The distance covered in 2 s is 18 m.

(b) is correct.

Explanation:

Given that,

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

Time t = 2 sec

The acceleration is the rate of change of velocity of the object,

So, $a = \dfrac{dv}{dt}$

$dv=(6t+5)dt$

On integrating w.r.to t

$\int_{0}^{v}v=\int_{0}^{t}(6t+5)dt$

$v = 3t^2+5t+C$

Where, C is a constant

Now, v = o and t = 0 then C = 0

$v = 3t^2+5t$

The velocity is the first derivative of the position of the object.

$v = \dfrac{dx}{dt}$

$dx=(3t^2+5t)dt$

On integrating within the conditions of motion as t  changes 0 to 2 and x changes 0 to x.

$\int_{0}^{x}x=\int_{0}^{2}(3t^2+5t)dt$

$x= [t^3+\dfrac{5t^2}{2}]_{0}^{2}$

$x=8+10$

$x = 18\ m$

Hence, The distance covered in 2 s is 18 m.