If an object with uniform acceleration (or deceleration) has a speed of #4 m/s# at #t=2# and moves a total of 25 m by #t=6#, what was the object's rate of acceleration?

Question

Charles Aeschliman · Accepted Answer

The acceleration is #=1.125ms^-2#

The initial speed is #u=4ms^-1#

The distace is #s=25m#

The time is #Deltat=6-2=4s#

Apply the equation of motion

#s=uDeltat+1/2a(Deltat)^2#

The acceleration is

#a=(s-uDeltat)/(1/2(Deltat)^2)#

#=(25-4xx4)/(1/2xx4^2)#

#=(9)/(8)#

#=1.125ms^-#

Olivia Askew · Answer

undefined To find the object's rate of acceleration, we can use the following formula:

$$v_f = v_i + at$$

Given: 
- $v_i = 4 \, \text{m/s}$ (initial velocity)
- $v_f = ?$ (final velocity)
- $t = 6 - 2 = 4 \, \text{s}$ (time interval)
- $d = 25 \, \text{m}$ (total distance traveled)

Using the formula:

$$v_f = v_i + at$$

$$v_f = 4 \, \text{m/s} + a(4 \, \text{s})$$

$$v_f = 4 \, \text{m/s} + 4a$$

Now, we also know that:

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

Plugging in the given values:

$$25 \, \text{m} = (4 \, \text{m/s})(4 \, \text{s}) + \frac{1}{2}a(4 \, \text{s})^2$$

$$25 \, \text{m} = 16 \, \text{m} + 8a$$

$$8a = 25 \, \text{m} - 16 \, \text{m}$$

$$8a = 9 \, \text{m}$$

$$a = \frac{9 \, \text{m}}{8 \, \text{s}^2}$$

$$a = 1.125 \, \text{m/s}^2$$

So, the object's rate of acceleration is $1.125 \, \text{m/s}^2$.