Question

An automobile starts from rest and accelerates to a final velocity in two stages along a straight road. Each stage occupies the same amount of time. In stage 1, the magnitude of the car's acceleration is 2.8 m/s2. The magnitude of the car's velocity at the end of stage 2 is 2.6 times greater than it is at the end of stage 1. Find the magnitude of the acceleration in stage 2.

Answer 1

The answer is a2 = 4.48 m/s2

Step-by-step explanation:

Let´s start with the equations we know:

1. 
   `v_(f) = v_(o) +at`
2. 
   `t_(1) =t_(2) =t`

Where:

• 
  `v_(f)` => Final velocity

• 
  `v_(o)` => Initial velocity
• a => acceleration
• t => time

Now, let´s divide the problem in Stage 1 and 2 and get equations for each stage:

Stage 1 knowns and unknowns:

• 
  `v_(1o) =0\\v_(1f) = ?\\a_(1) =2.8[(m)/(s^(2) )] \\t_(1) =?`

Stage 1 equation:

• 
  `v_(1f) = v_(1o) + a_(1) t`
• 
  `v_(1f)= 0+2.8t\\v_(1f)=2.8t`

Stage 2 knowns and unknowns:

• 
  `v_(2o) =v_(1f) \\v_(2f)=2.6v_(1f) \\a_(2) =?\\t_(2)=?`

Stage 2 equation:

• 
  `v_(2f) = v_(2o) +a_(2)t`
• 
  `2.6v_(1f) =v_(1f) +a_(2)t\\2.6v_(1f) -v_(1f) = a_(2)t\\1.6v_(1f) = a_(2)t`

Now we can substitute the resultant equation from stage 1 into stage´s 2 equation:

• 
  `1.6(2.8t) = a_(2)t`

We can see "t" is on both sides, so it cancels out and we are left with:

`a_(2) = 4.48 [(m)/(s^(2) ) ]`