Question

Kinematics in 1D [4B.R]:Question 6

An object with an initial velocity of 3.50 m/s moves east
along a straight and level path. The object then
undergoes a constant acceleration of 1.80 m/s^2 east for
a period of 5.00 s. How far does the object move while
it is accelerating?
Select one:
OC) 27.2 m
OA) 6.30 m
OB) 17.5 m
OD) 40.0 m

Answer 1

Using the kinematic equation s = ut + ½ at², the object's displacement over 5 seconds with an initial velocity of 3.50 m/s and constant acceleration of 1.80 m/s² east is calculated to be 40.0 meters.

Step-by-step explanation:

To calculate the displacement of an object with initial velocity and constant acceleration, we can use the kinematic equation:

s = ut + ½ at²

where:

• s is the displacement
• u is the initial velocity
• a is the acceleration
• t is the time

Substitute the given values into the equation to find the displacement:

s = (3.50 m/s)(5.00 s) + ½ (1.80 m/s²)(5.00 s)²

= 17.5 m + ½ (1.80 m/s²)(25 s²)

= 17.5 m + 0.9 m/s² × 25 s²

= 17.5 m + 22.5 m

= 40.0 m

Therefore, the object moves 40.0 meters while it is accelerating.