Question

An object has an acceleration of +1.2cm/s². At t=4.0s, its velocity is −3.4cm/s. Determine the object’s velocities at t=1.0s and t=6.0s.

a) 1.8cm/s, 5.0cm/s
b) 0.2cm/s, -6.8cm/s
c) -2.2cm/s, -4.2cm/s
d)-0.6cm/s, 7.2cm/s

Answer 1

The object's velocities at t = 1.0 s and t = 6.0 s are not provided in the given information.

Step-by-step explanation:

To determine the object's velocities at t = 1.0 s and t = 6.0 s, we can use the kinematic equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time interval.

At t = 1.0 s, the initial velocity (u) is not given, so we cannot determine the velocity at that time.

At t = 6.0 s, the initial velocity (u) is given as -3.4 cm/s and the acceleration (a) is given as +1.2 cm/s². Plugging these values into the kinematic equation, we get v = -3.4 cm/s + (1.2 cm/s²)(6.0 s) = -3.4 cm/s + 7.2 cm/s = 3.8 cm/s.

Therefore, the object's velocities at t = 1.0 s and t = 6.0 s are not provided in the given information.