Distance traversed by a particle moving along a straight line is given by [180t + 50t^2]m. Find

1. Break Down the Problem

We are given the distance $s(t)$ of a particle over time, which is represented by the equation: $s(t) = 180t + 50t^2$ We need to derive several parameters concerning the motion of this particle, such as velocity and acceleration.

2. Relevant Concepts

1. Velocity is the first derivative of distance with respect to time: $v(t) = \frac{ds}{dt}$
2. Acceleration is the second derivative of distance or the first derivative of velocity: $a(t) = \frac{d^2s}{dt^2} \quad \text{or} \quad a(t) = \frac{dv}{dt}$

3. Analysis and Detail

1. Finding Velocity: $v(t) = \frac{ds}{dt} = \frac{d}{dt}(180t + 50t^2)$ Applying the power rule: $v(t) = 180 + 100t$

2. Finding Acceleration: $a(t) = \frac{dv}{dt} = \frac{d}{dt}(180 + 100t)$ Again applying the power rule: $a(t) = 100$ This indicates that the acceleration is constant.

4. Verify and Summarize

• The velocity function $v(t) = 180 + 100t$ shows that the particle's speed increases linearly over time due to the constant acceleration.
• The calculation for acceleration $a(t) = 100$ confirms that the particle experiences a uniform acceleration.

Final Answer

1. Velocity Function: $v(t) = 180 + 100t$ m/s
2. Constant Acceleration: $a(t) = 100$ m/s²