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Step-by-Step Solution
Step 1: Split the motion into two phases
The car accelerates from rest with a constant acceleration $ \alpha $ for a certain time $ t_1 $ and then decelerates with a constant deceleration $ \beta $ for the remaining time $ t_2 $, until it comes to rest again. Let:
$ t_1 $ = time spent accelerating
$ t_2 = t - t_1 $ = time spent decelerating
Step 2: Express the velocity at the end of the acceleration phase
Since the car starts from rest and accelerates at $ \alpha $ for time $ t_1 $, its velocity at the end of this phase is:
$ v_1 = 0 + \alpha t_1 = \alpha t_1 $
Step 3: Relate velocity to deceleration phase
During deceleration, the car comes to rest again at the end of time $ t_2 = t - t_1 $. Starting this phase with initial velocity $ v_1 $ and decelerating at rate $ \beta $, the final velocity must be zero:
$ 0 = v_1 - \beta \, t_2 \\
v_1 = \beta \, t_2 = \beta ( t - t_1 )
$
But we already have $ v_1 = \alpha t_1 $. Equating these expressions for $ v_1 $ gives:
$ \alpha t_1 = \beta ( t - t_1 ) \\
\alpha t_1 = \beta t - \beta t_1 \\
\alpha t_1 + \beta t_1 = \beta t \\
t_1 (\alpha + \beta) = \beta t \\
t_1 = \frac{\beta \, t}{\alpha + \beta}
$
Step 4: Calculate distance traveled during acceleration ($ s_1 $)
While accelerating from rest with constant acceleration $ \alpha $ for time $ t_1 $, the distance covered is:
$ s_1 = \frac{1}{2} \alpha t_1^2
$
Step 5: Calculate distance traveled during deceleration ($ s_2 $)
During the deceleration phase, the initial velocity is $ v_1 = \alpha t_1 $ and the final velocity is zero. The time for deceleration is $ t_2 = t - t_1 $. Using the average-velocity formula for uniformly accelerated motion, the distance covered is:
$ s_2 = \frac{1}{2} \bigl(\text{initial velocity} + \text{final velocity}\bigr) \times \bigl(\text{time interval}\bigr)
= \frac{1}{2} \left(\alpha t_1 + 0\right) (t - t_1)
= \frac{\alpha t_1}{2} (t - t_1)
$
Step 6: Combine distances and simplify
The total distance traveled, $ s $, is:
$
s = s_1 + s_2
= \frac{1}{2} \alpha t_1^2 + \frac{\alpha t_1}{2} (t - t_1)
$
Factor out $ \frac{\alpha t_1}{2} $:
$
s = \frac{\alpha}{2} \Bigl( t_1^2 + t_1(t - t_1) \Bigr)
= \frac{\alpha}{2} \Bigl( t_1^2 + t_1 t - t_1^2 \Bigr)
= \frac{\alpha}{2} \, t_1 \, t
$
Since we found $ t_1 = \frac{\beta \, t}{\alpha + \beta} $, substitute this value into the expression for $ s $:
$
s = \frac{\alpha}{2} \left(\frac{\beta \, t}{\alpha + \beta}\right) t
= \frac{\alpha \beta}{2(\alpha + \beta)} \, t^2
$
Step 7: Final Answer
$ \displaystyle \text{Total distance traveled, } s = \frac{\alpha \beta}{2 (\alpha + \beta)}\, t^2.
$
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