JEE Main & Advanced Physics Motion in a Straight Line / सरल रेखा में गति Question Bank Critical Thinking Questions

  • question_answer
    A car accelerates from rest at a constant rate \[\alpha \]for some time, after which it decelerates at a constant rate \[\beta \]and comes to rest. If the total time elapsed is t, then the maximum velocity acquired by the car is  [IIT 1978; CBSE PMT 1994]

    A)             \[\left( \frac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \beta } \right)\,t\]

    B)               \[\left( \frac{{{\alpha }^{2}}-{{\beta }^{2}}}{\alpha \beta } \right)\,t\]

    C)             \[\frac{(\alpha +\beta )\,t}{\alpha \beta }\]

    D)               \[\frac{\alpha \beta \,t}{\alpha +\beta }\]

    Correct Answer: D

    Solution :

                   Let the car accelerate at rate \[\alpha \] for time \[{{t}_{1}}\] then maximum velocity attained, \[v=0+\alpha {{t}_{1}}=\alpha {{t}_{1}}\]             Now, the car decelerates at a rate \[\beta \] for time \[(t-{{t}_{1}})\] and finally comes to rest. Then,              \[0=v-\beta (t-{{t}_{1}})\]Þ \[0=\alpha {{t}_{1}}-\beta t+\beta {{t}_{1}}\]             Þ \[{{t}_{1}}=\frac{\beta }{\alpha +\beta }t\]             \ \[v=\frac{\alpha \beta }{\alpha +\beta }t\]


You need to login to perform this action.
You will be redirected in 3 sec spinner