Question

A particle moves on x -axis as per equation x=(t^(3)-9t^(2)+15t+2)m .Distance travelled by the particle between t=0 and t=5 s is (A) 25m (B) 39m (C) 23m (D) 52m​

Answer 1

Answer:

answer is 39 m

Explanation:

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Answer 2

Answer:

The magnitude of the distance travelled by the particle from t = 0s to t = 5s is 25m.

Explanation:

Given the distance travelled by a particle on x-axis is,

$x=\big(t^3-9t^2+15t+2\big)m$

To find the distance travelled at time, t = 0s, substitute t = 0 in the above equation,

$x(0)=0^3-9\times 0^2+15\times 0+2=2m$

To find the distance travelled at time, t = 5s, substitute t = 5 in the above equation,

$x(5)=5^3-9\times 5^2+15\times 5+2$

       $=125-225+75+2=-23m$

Thus, the distance travelled by the particle from t = 0s to t = 5s is

$x(5)-x(0)$

$=-23-2=-25m$

The distance travelled by the particle is -25m.

The minus sign shows the negative direction.  

Therefore, the magnitude of the distance travelled by the particle from t = 0s to t = 5s is 25m.

So, the correct answer is option A.

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