Question

A car accelerates from rest at a constant rate α for some time after which it decelerates at a constant rate β to come to rest. If the total time elapsed is t, the maximum velocity acquired by the car is given by: 

1. \(\left(\dfrac{\alpha^2 + \beta^2}{\alpha \beta}\right)t\)
2. \(\left(\dfrac{\alpha^2 - \beta^2}{\alpha \beta}\right)t\)
3. \(\left(\dfrac{\alpha + \beta}{\alpha \beta}\right)t\)
4. \(\left(\dfrac{\alpha\beta}{\alpha +\beta}\right)t\)

Answer 1

Correct Answer - Option 4 : \(\left(\dfrac{\alpha\beta}{\alpha +\beta}\right)t\)

Concept:

• Acceleration: The rate of change of velocity with respect to time is called acceleration.

\(a = \frac{dv}{dt}\)

• Deceleration: The acceleration in which speed is being reduced is termed as deceleration. It is negative of acceleration.

For a uniformly accelerated motion, the equations of motion is represented as 

v = u + at -- (1)

\(S = ut + \frac{1}{2}at^2\) -- (2)

v ² = u ² + 2as -- (3)

Where v is final velocity, u is initial velocity, t is the time interval for which the force is applied, s is displacement.

Calculation:

The body is accelerated for a while with acceleration α and then decelerated with acceleration β 

The highest velocity will be the point where the object is accelerated to its peak value of velocity, and the deacceleration is about to start.

Let body is accelerated for time t₁ and then deaccelerated for time t₂

total time taken is t 

So, 

t₁ + t₂ = t -- (4)

Let maximum velocity is v, which is obtained after being accelerated for α for t₁ seconds. 

The body started from rest so, initial velocity u = 0

Now, using this information in the first equation of motion

v = 0 +  α t1 

⇒ v = α t1 --- (5)

Now, the body is deaccelerated with β  from v till the final velocity becomes zero in time t₂

0 = v - β t2 ---- (6)

⇒ Using (5) and (6) we get 

α t1 = β t2 

 \(\implies \frac{t_1}{t_2} = \frac{\beta}{\alpha }\)

\(\implies t_2 = (\frac{\alpha }{\beta}) t_1\) -- (7)

Putting (7) in (4)

\(t= t_1 + (\frac{\alpha }{\beta}) t_1\)

\(\implies t= t_1 (1 + \frac{\alpha}{\beta})\)

\(\implies t_1 = \frac{\beta }{\alpha + \beta} t\) --- (8)

Putting (8) in (5) we get

\(v = \left(\dfrac{\alpha\beta}{\alpha +\beta}\right)t\)

So, the correct answer is \(\left(\dfrac{\alpha\beta}{\alpha +\beta}\right)t\)