Question

An object moves with constant acceleration 4.05 m/s2 and over a time interval reaches a final velocity of 11.2 m/s.
(a) If its original velocity is 5.60 m/s, what is its displacement during the time interval?
m

(b) What is the distance it travels during this interval?
m

(c) If its initial velocity is
−5.60 m/s,
what is its displacement during this interval?
m

(d) What is the total distance it travels during the interval in part (c)?
m

Answer 1

a) Displacement: 11.6 m

b) Distance: 11.6 m

c) Displacement: 11.6 m

d) Distance: 19.4 m

Explanation:

a)

The motion of the object is a uniformly accelerated motion, so we can use the following suvat equation to find the displacement:

$v^2-u^2=2as$

where

s is the displacement

u is the initial velocity

a is the acceleration

v is the final velocity

For the object in this problem,

u = 5.60 m/s

v = 11.2 m/s

$a=4.05 m/s^2$

Solving for s, we find the displacement:

$s=\frac{v^2-u^2}{2a}=\frac{11.2^2-5.60^2}{2(4.05)}=11.6 m$

b)

Since the motion of the object is in a straight line and the direction of the velocity does not change, the distance is equal to the displacement, therefore

d = 11.6 m

c)

We can find the displacement of the object in this case by using again the equation

$v^2-u^2=2as$

where

s is the displacement

u is the initial velocity

a is the acceleration

v is the final velocity

This time, we have:

u = -5.60 m/s

v = 11.2 m/s

$a=4.05 m/s^2$

Solving for s, we find the displacement:

$s=\frac{v^2-u^2}{2a}=\frac{11.2^2-(-5.60)^2}{2(4.05)}=11.6 m$

d)

In this case, the velocity of the object changes direction: this means that for part of the trip the object moves in the negative direction, and for part of the trip it moves in the positive direction. Therefore, the distance is not equal to the displacement.

In order to find the distance, we have to calculate first the distance covered when the object is moving in the negative direction; so, from when it has velocity of

u = -5.60 m/s

to when it has a velocity of

v = 0

Therefore,

$d_1=\frac{v^2-u^2}{2a}=\frac{0-(-5.60)^2}{2(4.05)}=3.9 m$

Then, we have to add the distance covered from when the object has velocity

u = 0

to when it has velocity of

v = 11.2 m/s

Therefore,

$d_2=\frac{v^2-u^2}{2a}=\frac{11.2^2-(0)^2}{2(4.05)}=15.5 m$

Therefore, the total distance covered is

$d=d_1+d_2=3.9 + 15.5=19.4 m$

Learn more about distance and displacement:

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