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

Answer:

The answer is a2 = 4.48 m/s2

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:

• $v_{1o} =0\\v_{1f} = ?\\a_{1} =2.8[\frac{m}{s^{2} }] \\t_{1} =?$

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:

$a_{2} = 4.48 [\frac{m}{s^{2} } ]$