A car accelerates from rest at a constant rate α for some time after which it decelerates at a constant…

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 seconds, the total distance travelled is:
  1. 4αβ(α+β)t2
  2. 2αβ(α+β)t2
  3. αβ2(α+β)t2
  4. αβ4(α+β)t2

Solution

v0=αt1 and 0=v0-βt2⇒v0=βt2

t1+t2=t

v01α+1β=t

⇒v0=αβtα+β

Distance = area of v-t graph

=12×t×v0=12×t×αβtα+β=αβt22(α+β)

Asked in: JEE Main 2021 (17 Mar Shift 1)

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