What Is the Minimum Time for a Sportscar to Travel 1/2 Mile from Standstill?

[ttja]

Homework Statement

a A sportscar can accelerate uniformly to 120 mi/h in 30s. Its maximum braking rate cannot exceed 0.7g. what is the minimum time required to go 1/2 mi, assuming it begins and ends at rest?

Homework Equations

I drew a graph of v(t) vs t. where the initial acc. goes up to a certain time, t1, then decelerate to a rest at t2.

The Attempt at a Solution

acc= 4 mi/min
V1 = 4t

For convenience i drew another graph for when the car dec. Therefore:

acc= -0.7g
V2 = 4t1 - 0.7gt

Integrate:

d = 0.5 = int ( 4t, t, 0, t1) + int (4t1 - 0.7gt, t, 0, t2)

=> 2t1^2 + 4t1t2 - 0.7g/2 t2^2

Here is where i got lost. i know that t = t1 + t2 but i don't know how to derive t with its components to get the minimum time. please help and thank you.

 

[hotvette]

Interesting problem. You can determine t₂ in terms of t₁ because the velocity must be zero in the end. Other considerations are less than max deceleration and coast time between acceleration and deceleration.

 

[tomcue]

I would approach this problem by first identifying the key variables and equations involved. The key variables in this problem are the acceleration (a), velocity (v), and displacement (d), as well as the initial and final times (t1 and t2). The relevant equations are the equations for constant acceleration, which are v = v0 + at and d = v0t + 1/2at^2.

Using the given information, we can determine that the car's acceleration is 4 mi/min or 0.067 mi/s^2, and its maximum braking rate is 0.7g or 6.87 mi/s^2. We also know that the initial velocity is 0 mi/h and the final velocity is 120 mi/h.

To find the minimum time required to go 1/2 mi, we can set up the following equation using the equations for constant acceleration:

1/2 = (120 mi/h)t2 + 1/2(0.067 mi/s^2)(t2)^2

Solving for t2, we get t2 = 2.5 s.

Next, we can use the same equation to find t1, the time it takes for the car to accelerate to its maximum velocity of 120 mi/h.

1/2 = (0 mi/h)t1 + 1/2(0.067 mi/s^2)(t1)^2

Solving for t1, we get t1 = 30 s.

Therefore, the minimum time required to go 1/2 mi is t = t1 + t2 = 32.5 s.

In conclusion, the key to solving this problem is to set up and solve the relevant equations for constant acceleration, while keeping in mind the constraints given (maximum braking rate and initial and final velocities).