A car accelerates from rest at a constant rate α forsome time after w

1.	A car accelerates from rest at a constant rate α forsome time after which it decelerates at a constantrate β to come to rest. If the total time is t provethat maximum velocity reached is Vm=αßt/α+β and total displacement is s=αßt²/2(α + β)
Answer» For the first leg of the journey, V = 0 + α t1So V = α t1 for the second leg of journey of duration t - t1 :0 = V - β (t - t1)0 = α t1 - β t + β t1 t1 = β t / (α + β) V = αt1 = α β t / (α + β ) Let ‘t1’ be the time for which it accelerates, ‘v’ be the velocity after this time. Now, when the acceleration is α. v = 0 + αt1 => v = αt1 => t1= v/α Then it decelerates at rate β and stops at t2. So, 0 = v – βt2 => v = βt2 => t2= v/β Now, t = t1+ t2= [v/α] + [v/β] = v[(α+β)/αβ] => v = t[αβ/(α+β)] Now, Distance travelled in time t1is found as, v2= 02+ 2αS1 => S1= v2/(2α) Distance travelled in time t1is found as, 02= v2- 2βS2 => S2= v2/(2β) Total distance travelled is, S = S1+ S2 => S = v2/(2α) + v2/(2β) => S = [v2/2][1/α + 1/β] => S = [v2/2] [(α+β)/αβ] => S = ½ [t[αβ/(α+β)]]2[(α+β)/αβ] => S = ½ t[αβ/(α+β)]

A car accelerates from rest at a constant rate α forsome time after which it decelerates at a constantrate β to come to rest. If the total time is t provethat maximum velocity reached is Vm=αßt/α+β and total displacement is s=αßt²/2(α + β)

Answer»

For the first leg of the journey, V = 0 + α t1So V = α t1

for the second leg of journey of duration t - t1 :0 = V - β (t - t1)0 = α t1 - β t + β t1

t1 = β t / (α + β)

V = αt1 = α β t / (α + β )

Let ‘t1’ be the time for which it accelerates, ‘v’ be the velocity after this time. Now, when the acceleration is α.

v = 0 + αt1

=> v = αt1

=> t1= v/α

Then it decelerates at rate β and stops at t2.

So,

0 = v – βt2

=> v = βt2

=> t2= v/β

Now, t = t1+ t2= [v/α] + [v/β] = v[(α+β)/αβ]

=> v = t[αβ/(α+β)]

Now,

Distance travelled in time t1is found as,

v2= 02+ 2αS1

=> S1= v2/(2α)

Distance travelled in time t1is found as,

02= v2- 2βS2

=> S2= v2/(2β)

Total distance travelled is,

S = S1+ S2

=> S = v2/(2α) + v2/(2β)

=> S = [v2/2][1/α + 1/β]

=> S = [v2/2] [(α+β)/αβ]

=> S = ½ [t[αβ/(α+β)]]2[(α+β)/αβ]

=> S = ½ t[αβ/(α+β)]