Question

An object starts at Xi = -4m with an initial velocity of 4m/s. It experiences an acceleration of -2 m/s^2 for 2 seconds, followed by an acceleration of -6 m/s^2 for 4 seconds. 1. After 2 seconds, what is the objects velocity? 2. After 6 seconds, what is the objects velocity? 3. After 6 seconds, what is the objects total displacement?

Answer 1

Answer:

a) 0 m/s

b) - 24 m/s

c)  - 68 m

Explanation:

Given:

Initial distance = - 4 m

Initial velocity, u = 4 m/s

1) acceleration, a = - 2 m/s² for time, t = 2 seconds

thus,

velocity after 2 seconds will be

from Newton's equation of motion

v = u + at

v = 4 + (-2) × 2

v = 0 m/s

2) Velocity after 2 second is the initial velocity for this case

given acceleration = - 6 m/s² for 4 seconds

thus,

final velocity, v = 0 + ( - 6 ) × 4 = - 24 m/s

here the negative sign depicts the velocity in opposite direction to the initial direction of motion

thus, velocity after 6 seconds = - 24 m/s

3) Now,

Total displacement in 6 seconds

= Displacement in 2 seconds + Displacement in 4 seconds

From Newton's equation of motion

$s=ut+\frac{1}{2}at^2$

where,  

s is the distance

u is the initial speed  

a is the acceleration

t is the time

thus,

= $0\times2+\frac{1}{2}\times(-2)\times2^2$  + $0\times4+\frac{1}{2}\times-6\times4^2$

= - 16 - 48

= - 64 m

Hence, the final displacement = - 64 - 4 = - 68 m