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Motion in a Straight Line — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsMotion in a Straight Line5 Marks
Question
A car accelerates from rest at a constant rate α for some time, after which it decelerates at a constant rate $\beta$ to come to rest. If t is the total time elapsed, then calculate.
  1. The maximum velocity attained by the car.
  2. The total distance travelled by the car.
✓

Answer

  1. Let the car accelerate at a rate $\alpha$ for time $t_1$ and attain a maximum velocity $v$. Then the car decelerates at a constant rate $\beta$ for remaining time $(t - t_1)$ and again comes to rest. Then from adjoining figure, it is clear that

$\text{v}=\alpha\text{t}_1=\beta(\text{t}-\text{t}_1)$
$\therefore \text{t}_1=\frac{\text{v}}{\alpha}$
$(\text{t}-\text{t}_1)=\frac{\text{v}}{\beta}$
Adding these two, we have
$\text{t}=\frac{\text{v}}{\alpha}+\frac{\text{v}}{\beta}=\text{v}\Big(\frac{\alpha+\beta}{\alpha\beta}\Big)$
$\Rightarrow \text{v}=\frac{\alpha\beta\text{t}}{(\alpha+\beta)}$
  1. Total distance travelled by the car $s =$ area $\text{OAB} =\frac{1}{2}(\text{t})\times(\text{v})$
$\therefore \text{s}=\frac{1}{2}\text{t}\times\frac{\alpha\beta\text{t}}{(\alpha+\beta)}$
$=\frac{1}{2}\frac{\alpha\beta}{(\alpha+\beta)}\text{t}^2$

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