Question

Consider motion in one dimension. (The sign of the vector quantities is their direction indicator.) An object moves in the positive x-direction with speed 0.9 m/s for 3.6 s. It stops for 4.7 s and then moves in the negative x direction with speed 1 m/s for 2.9 s. What is the total distance traveled by the object in units of meter? Enter a number with two digits behind the decimal point.

Answer 1

6.14 m

Step-by-step explanation:

The motion of the object is divided into three parts:

- Part 1: the object moves with speed

v = +0.9 m/s (the positive sign means positive x-direction)

for t = 3.6 s

Therefore, the distance covered is

`d_1 =vt=(0.9 m/s)(3.6 s)=3.24 m`

- Part 2: the object stops for 4.7 s, so during this time the distance travelled by the object is zero.

- Part 3: the object moves with speed

v = -1 m/s (negative sign means negative x-direction)

for a time of

t = 2.9 s

We are only concerned in the distance travelled, so we can ignore the negative sign, and the distance covered in this part is

`d_3 = vt=(1 m/s)(2.9 s)=2.90 m`

So, the total distance covered is

`d=d_1+d_2+d_3=3.24 m+0 +2.90 m=6.14 m`