Question

A particle moves along one dimension with a constant acceleration of 4.60 m/s2 over a time interval. At the end of this interval it has reached a velocity of 13.8 m/s.
(a)
If its original velocity is 6.90 m/s, what is its displacement (in m) during the time interval?
m
(b)
What is the distance it travels (in m) during this interval?
m
(c)
A second particle moves in one dimension, also with a constant acceleration of 4.60 m/s2 , but over some different time interval. Like the first particle, its velocity at the end of the interval is 13.8 m/s, but its initial velocity is
−6.90 m/s.
What is the displacement (in m) of the second particle over this interval?
m
(d)
What is the total distance traveled (in m) by the second particle in part (c), during the interval in part (c)?
m

Answer 1

Answer:

Part a)

$d = 15.53 m$

Part b)

$d = 15.53 m$

Part c)

$d = 15.53 m$

Part d)

$d = 25.87 m$

Explanation:

As we know that

initial velocity is 6.90 m/s

Acceleration is given as

$a = 4.60 m/s^2$

final velocity is 13.8 m/s

Part a)

displacement is given as

$v_f^2 - v_i^2 = 2 a d$

$13.8^2 - 6.90^2 = 2 (4.60)d$

$d = 15.53 m$

Part b)

Since it is a one dimensional motion so distance and displacement must be same

$d = 15.53 m$

Part c)

displacement is given as

$v_f^2 - v_i^2 = 2 a d$

$13.8^2 - (-6.90)^2 = 2(4.60)d$

$d = 15.53 m$

Part d)

now the displacement when it stop

$0 - (-6.90)^2 = 2(4.60)d$

$d_1 = -5.175 m$

Now from this it will reach to its final speed 13.8 m/s so we have

$13.8^2 - 0 = 2(4.60)d_2$

$d_2 = 20.7 m$

so total distance moved is

$d = 20.7 + 5.17$

$d = 25.87 m$