Question

An object with an initial velocity of 3.50 m/s moves east along a straight level path the object then undergoes a constant acceleration of 1.80 m/s east for a period of 5.00 s. How far does the object move while it is accelerating

Answer 1

distance traveled by the object is given as

$x = v_i t + \frac{1}{2} at^2$

given that

$v_i = 3.50 m/s$

$a = 1.80 m/s^2$

$t = 5 s$

now plug in all the values above

$x = (3.50)(5) + \frac{1}{2}(1.80)(5)^2$

$x = 40 m$

so here it will move total distance of 40 m

Answer 2

The distance covered by the object while it is accelerating is 40 meters.

Given the following data:

• Initial velocity = 3.50 m/s.

• Acceleration = 1.80 $m/s^2$

• Time = 5.00 seconds.

To determine the distance covered by the object while it is accelerating, we would apply the second equation of motion:

The second equation of motion.

Mathematically, the second equation of motion is given by this formula:

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

Where:

• S is the distance.

• u is the initial velocity.

• a is the acceleration.

• t is the time.

Substituting the given parameters into the formula, we have;

$S=3.5(5)+\frac{1}{2} \times 1.80 \times 5^2\\\\S=17.5+0.9 \times 25\\\\S=17.5+22.5$

S = 40 meters.

Read more on acceleration here: brainly.com/question/24728358