PhysicsMedium65×since 2002Q7847A car accelerates from rest at a constant rate α\alphaα for some time after which it decelerates at a constant rate β\betaβ to come to rest. If the total time elapsed is t seconds, the total distance travelled is :A4αβ(α+β)t2{{4\alpha \beta } \over {(\alpha + \beta )}}{t^2}(α+β)4αβt2B2αβ(α+β)t2{{2\alpha \beta } \over {(\alpha + \beta )}}{t^2}(α+β)2αβt2Cαβ2(α+β)t2{{\alpha \beta } \over {2(\alpha + \beta )}}{t^2}2(α+β)αβt2Dαβ4(α+β)t2{{\alpha \beta } \over {4(\alpha + \beta )}}{t^2}4(α+β)αβt2Check answerSkip