Question

A car accelerates from rest at a constant rate alpha for sometime after which it decelerates at a constant rate beta to come to rest. If total time lapse be t, evaluate the 1) maximum velocity
reached
2) total distance traveled.

Answer 1

Let t₁ be time for which it accelerates, v be velocity after t₁,

Now, when the acceleration is α

v = 0 + α t₁

v = α t₁

t₁ = v/α

Then it decelerates at rate β, and stops at t₂

0 = v - β t₂

Initial velocity before t₂ = v = α t₁

v = β t₂

α t₁ = β t₂

t₂ = (α/β) t₁

t = t₁ + t₂ 

t = t₁ + (α/β) t₁ = t₁ (1 + α/β)

or  t₁ = t / (1 + α/β)

So, max velocity attained = α t₁

= α t / (1 + α/β) = t [ 1/α + 1/β ]

Also we can write

t₂ = v/β

Now, t = t₁ + t₂

t = v/α + v/β

t = v [ (α + β) / αβ ]

Distance in t₁ is given by

v² = 0² + 2 α s₁

s₁ = v² / 2α

Distance in t₂ is given by

0² = v² - 2 β s₂

s₂ = v² / 2β

S = s₁ + s₂

So, S = v² / 2α + v² / 2β

S = v²/2 [ 1/α + 1/β ]

S = v²/2 [ (α + β)/αβ ]

S = 1/2 [ t { αβ/(α + β) } ]² [ (α + β)/αβ ]

S = 1/2 [ t { αβ/(α + β) } ]

Answer 2

Let total time taken is t.
And t₁ is time taken in accelerating motion .
∴ (t - t₁) is the time taken in decelerating motion .

Now, use formula,
v = u + at
For accelerating,
u = 0, a = α, time = t₁
∴ v = 0 + αt₁
v = αt₁ --------(1)

For decelerating motion,
v = 0, a = -β , time = (t - t₁)
∴ 0 = u - β(t - t₁)
u = βt - βt₁-----------(2)

because car first accelerates and then decelerates . So,
Final velocity in first case (accelerating) = initial velocity in 2nd case(decelerating)= maximum velocity
e.g., v = u

From equation (1) and (2),
v = βt - vβ/α
vα = αβt - vβ
v(α + β) = αβt
v = αβt/(α + β)
Hence, maximum velocity = αβt/(α + β)

Now, totol distance covered by car = distance covered during acceleration+ distance covered during deceleration
=1/2αt₁² + [αβt/(α + β) × t - 1/2β(t - t₁)² ]
= 1/2α v²/α² + [αβt²/(α + β) - 1/2β × u²/β² ]
= αβ²t²/2(α + β)² + α²βt²/2(α + β)we
= αβt²/2(α + β)
Hence tortal distance covered = 1/2 αβ/(α + β) t²