A car starts from rest and travels with unifo | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $t_1$ be the time taken to accelerate at rate $\alpha$ to reach maximum velocity $v_{\max}$,and $t_2$ be the time taken to retard at rate $\be

Vedclass · Accepted Answer

$\frac{\alpha \beta}{(\alpha + \beta)} t$

Vedclass · Answer

(A) Let $t_1$ be the time taken to accelerate at rate $\alpha$ to reach maximum velocity $v_{\max}$,and $t_2$ be the time taken to retard at rate $\beta$ to come to rest.From the velocity-time graph,the slope of the acceleration phase is $\alpha = \frac{v_{\max}}{t_1}$,which gives $t_1 = \frac{v_{\max}}{\alpha}$.The slope of the retardation phase is $\beta = \frac{v_{\max}}{t_2}$,which gives $t_2 = \frac{v_{\max}}{\beta}$.The total time is $t = t_1 + t_2$.Substituting the expressions for $t_1$ and $t_2$: $t = \frac{v_{\max}}{\alpha} + \frac{v_{\max}}{\beta}$.$t = v_{\max} \left( \frac{\alpha + \beta}{\alpha \beta} \right)$.Solving for $v_{\max}$,we get $v_{\max} = \frac{\alpha \beta t}{\alpha + \beta}$.

$(1)$ Velocity decreases	$(a)$ Linear graph (constant slope)
$(2)$ Velocity increases	$(b)$ Concave up graph (increasing slope)
$(3)$ Velocity is constant	$(c)$ Concave down graph (decreasing slope)

$A$ car starts from rest and travels with uniform acceleration $\alpha$ for some time and then with uniform retardation $\beta$ and comes to rest. If the total travel time of the car is $t$,the maximum velocity attained by it is given by:

Similar Questions

Match Column-$I$ with Column-$II$ based on the distance-time $(d-t)$ graphs provided in the image.
$(1)$ Velocity decreases $(a)$ Linear graph (constant slope)
$(2)$ Velocity increases $(b)$ Concave up graph (increasing slope)
$(3)$ Velocity is constant $(c)$ Concave down graph (decreasing slope)

Two cars $P$ and $Q$ start from a point at the same time in a straight line and their positions are represented by $x_P(t) = at + bt^2$ and $x_Q(t) = ft - t^2$. At what time do the cars have the same velocity?

The position-time $(x-t)$ graph for positive acceleration is

For the given displacement-time graph,what will the corresponding velocity-time graph look like?

For the velocity-time graph shown in the figure,the distance covered by the body in the last two seconds of its motion is $S_1$. What is the ratio of $S_1$ to the total distance covered by it?

Vedclass Test Series

Exam Paper Generator

Online Exam Module