Question

A car starting from rest first moves with a acceleration of 5m/s2 for sometime and after moving with uniform speed starts retarding with same rate to come to rest in total time of 25 sec. If the average velocity of car for the whole journey is 20m/s, for how long does it move with a uniform speed.

Answer 1

@Luigi0210 can help you. I am sorry for not being of assistance. This is an extremely smart person.

Answer 2

@ApollosChariot

Answer 3

@Luigi0210

Answer 4

@No.name

Answer 5

@camerondoherty

Answer 6

First, we have to form a formula for the area under the graph. The speed at which the car travels uniformly is denoted by \(v_{0}\). Let x be the time needed to reach \(v_{0}\)m/s, the acceleration is
\[a=\frac{v_{0}}{x}\]
But we know a, which is 5m/s2. Rearranging, this gives:
\[v_{0}=5x\]
The area of the graph is the total distance travelled. I will skip ahead about getting the areas from the graph (triangles and rectange), but basically the areas give the right hand side of the equation, while the left hand side is the total distance travelled using \(distance=average\:speed\times total\:time\):
\[20\times25=\frac{1}{2}\times x \times v_{0} + v_{0}\times (25-2x) + \frac{1}{2}\times x \:\times\:v_{0}\]Substituting \(v_{0}=5x\) and solving for x should yield you the answers 20 or 5. At this point we cannot reject either answers.

Next, we have to find the time the car takes to travel at uniform speed. In the expression above the time is \(25-x\). Substituting x=20 gives -15s, which is to be rejected. Substituting x=5 gives 15s, which is what you are looking for.

Notes:
(1)The answer -15s is rejected as it is a solution to the case where the car does everything the same cept that it does it in reverse.
(2) The delayed rejection of answers can be avoided if the equations are formulated differently, if I am not mistaken this can be done with 25-2x=y.