If an object is moving at 8 m/s and accelerates to 32 m/s over 6 seconds, what was the object's rate of acceleration?

3 Answers
Jun 29, 2018

4 \ "m/s"^2

Explanation:

Well, the object changed its velocity by 32 \ "m/s"-8 \ "m/s" over 6 seconds.

Acceleration is given by the equation:

a=(s_f-s_i)/t

where:

  • s_f,s_i are the final and initial speeds, respectively

  • t is the time taken

So, we get:

a=(32 \ "m/s"-8 \ "m/s")/(6 \ "s")

=(24 \ "m/s")/(6 \ "s")

=4 \ "m/s"^2

Jun 29, 2018

The acceleration is =4ms^-2

Explanation:

The initial velocity is u=8ms^-1

The final velocity is v=32ms^-1

The time is t=6s

Apply the equation of motion

v=u+at

The acceleration is

a=(v-u)/t=(32-8)/6=24/6=4ms^-2

Jun 29, 2018

Rate of acceleration a = 6 m/ s^2

Explanation:

v = u + a t

u = 8 m/s, v = 32 m/s, t = 6 sec

a = (v - u) / t = (32 - 8) / 6 = 4 m / s^2