Question

A particle travels along a straight line with a velocity v = (12 - 3t) m/s, where t is in seconds. When t = 1 s, the particle is located 10 m to the left of the origin. Determine the acceleration when t = 4 s. the displacement from t = 0,t = 10 s, and the distance the particle travels during this time period. a 9 2. A sphere is fired downwards into a medium with an initial speed of 27 m/s. If it experiences a deceleration of a = (-6t) m/s2 ,where t is in seconds, determine the distance traveled before it stops.

Answer 1

Answer:

(1).The acceleration is -24 m/s²

The total distance is 912 m.

(2). The distance traveled before it stop is 78 m.

Explanation:

Given that,

Velocity = (12-3t^2) m/s

When t = 1 s, the particle is located 10 m to the left of the origin.

We need to calculate the acceleration at t = 4 sec

Using formula of acceleration

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

Put the value of v

$a=\dfrac{d}{dt}(12-3t^2)$

$a=-6t$

Put the value of t

$a=-6\times4$

$a=-24\ m/s^2$

The displacement from t = 0,t = 10 s, and the distance the particle travels during this time period.

We need to calculate the distance

Using formula of distance

$ds=v\ dt$

$\int_{-10}^{s}=\int_{1}^{t}{v}dt$

Put the value of v

$\int_{-10}^{s}=\int_{1}^{t}{ (12-3t^2)}dt$

$s+10=12t-t^3-11$

$s=12t-t^3-21$

At t = 0,

$s_{t=0}=-21$

At t = 10,

$s_{t=10}=12\times10-10^3-21$

$s_{t=10}=-901$

The displacement is

$\Delta s=-901-(-21)$

$\Delta s=-880\ m$

The distance at t= 2 sec

$s_{t=2}=12\times2-2^3-21$

$s_{t=2}=-5$

The total distance will be,

$s_{T}=(21-5)+(901-5)$

$s_{T}=912\ m$

(2). We need to calculate the distance at 2 sec

Using equation of motion

$s=ut-\dfrac{1}{2}at^2$

Put the value into the formula

$s=27\times 2+\dfrac{1}{2}\times6\times(2)^3$

$s=78\ m$

Hence, (1).The acceleration is -24 m/s²

The total distance is 912 m.

(2). The distance traveled before it stop is 78 m.