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:

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 seconds, the total distance travelled is: (A) (4 α β/(α+β)) t 2 (B) (2 α β/(α+β)) t 2 (C) (α β/2(α+β)) t 2 (D) (α β/4(α+β)) t 2

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/5d985f1438f2495c1e51df03ecb17557-.png" /> v 0=α t 1 and 0= v 0-β t 2 ⇒ v 0=β t 2 v 0((1/α)+(1/β))= t t 1+ t 2= t ⇒ v 0=(α β t /α+β) Distance = area of v - t graph =(1/2) × t × v 0=(1/2) × t × (α β t /α+β)=(α β t 2/2(α+β))

Q. 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:

$\frac{4 α β}{( α + β )} t^{2}$

$\frac{2 α β}{( α + β )} t^{2}$

$\frac{α β}{2 ( α + β )} t^{2}$

$\frac{α β}{4 ( α + β )} t^{2}$

Q. 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:

(α+β)4αβ​t2

(α+β)2αβ​t2

2(α+β)αβ​t2

4(α+β)αβ​t2

v0​=αt1​ and 0=v0​−βt2​ ⇒v0​=βt2​ v0​(α1​+β1​)=t t1​+t2​=t ⇒v0​=α+βαβt​ Distance = area of v−t graph =21​×t×v0​=21​×t×α+βαβt​=2(α+β)αβt2​

Q. 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:

$\frac{4 α β}{( α + β )} t^{2}$

$\frac{2 α β}{( α + β )} t^{2}$

$\frac{α β}{2 ( α + β )} t^{2}$

$\frac{α β}{4 ( α + β )} t^{2}$