Question

A particle, initially at rest, starts moving in a straight line with an acceleration a=6t+4m/s2. The
distance covered by it in 3 s is
(1) 30 m
(2) 60 m
(3) 45 m
(4) 15 m​

Answer 1

Answer:

45 m

Explanation:

i hope u will like my sol

Answer 2

Answer:

The distance covered by it in 3 s is 45 m.

(3) is correct option.

Explanation:

Given that,

Acceleration

$a = (6t+ 4) m/s^2$....(I)

Time t = 3 s

Suppose the integration constant is zero.

Acceleration :

The acceleration is the second derivative of the position of the particle.

We need to calculate the distance

$a =\dfrac{d^2x}{dt^2}$

$\dfrac{d^2x}{dt^2}=6t+4$

On integrating both side of equation (I)

$\int{\dfrac{d^2x}{dt^2}}=\int{6t+4}$

$\dfrac{dx}{dt}=3t^2+4t+C$

Here, C is integration constant

$\dfrac{dx}{dt}=3t^2+4t$

Again, On integrating

$\int{\dfrac{dx}{dt}}=\int{3t^2+4t}$

$x=t^3+2t^2+c'$

Here, C' is integration constant

The distance covered by it in 3 sec.

$x=3^3+2\times3^2$

$x=27+18= 45\ m$

Hence, The distance covered by it in 3 s is 45 m.