The acceleration `a` of a particle in `m/s^2` is give by `a=18t^2+36t+120`. If the particle starts from rest, then velocity at end of 1 s is

Options

A. 122 m/s
B. 140 m/s
C. 144 m/s
D. 120 m/s

Answer

To find the velocity of the particle at the end of 1 second, we need to integrate the given acceleration function with respect to time. The acceleration \( a \) is given by:

\[ a = 18t^2 + 36t + 120 \]

Since acceleration \( a \) is the derivative of velocity \( v \) with respect to time \( t \), we can write:

\[ a = \frac{dv}{dt} \]

Thus, we have:

\[ \frac{dv}{dt} = 18t^2 + 36t + 120 \]

To find the velocity, we will integrate both sides with respect to \( t \):

\[ \int dv = \int (18t^2 + 36t + 120) dt \]

Now, we perform the integration:

1. The integral of \( 18t^2 \) is \( 6t^3 \) (since \( \frac{18}{3} = 6 \)).
2. The integral of \( 36t \) is \( 18t^2 \) (since \( \frac{36}{2} = 18 \)).
3. The integral of \( 120 \) is \( 120t \).

Putting it all together, we have:

\[ v = 6t^3 + 18t^2 + 120t + C \]

Since the particle starts from rest, we know that the initial velocity \( v(0) = 0 \). We can use this to find the constant \( C \):

\[ v(0) = 6(0)^3 + 18(0)^2 + 120(0) + C = 0 \implies C = 0 \]

Thus, the velocity function simplifies to:

\[ v = 6t^3 + 18t^2 + 120t \]

Now, we need to find the velocity at \( t = 1 \) second:

\[ v(1) = 6(1)^3 + 18(1)^2 + 120(1) \]
\[ = 6(1) + 18(1) + 120 \]
\[ = 6 + 18 + 120 \]
\[ = 144 \, \text{m/s} \]

Therefore, the velocity of the particle at the end of 1 second is:

\[ \boxed{144 \, \text{m/s}} \]