a motorcycle starts from rest point A and travels 300m along a straight horizontal track to point B where it comes to a stop. if the acceleration is limited to 6.9m/s and deceleration to 5.9m/s calculate the least possible time to cover the distance and maximum velocity reached.

I Know that S1 + S2 = 300

but i dont know what formulas to use to replace S1 and S2 given the limited information:

S1:
a = 6.9m/s^2
u = 0
v = ?
t = ?
s = ?

S2
a= -5.9m/s^2
u = ?
v = 0
t = 0
s = ?

Overall distance = 300

Any help would be appreciated

Thanks

T = t₁ + t₂ since the minimum time requires full acceleration followed by full braking.

a₁ = +6.9 m/s²
a₂ = −5.9 m/s²

Since the initial and final velocity is the same, i.e. 0, we have
a₁ t₁ + a₂ t₂ = 0

or t₂ = −(a₁ t₁)/a₂
or T = (1 − a₁ /a₂) t₁

And since we know the total distance traveled is d = 300 meters, we write distance as a function of t₁ .

d = ½ a₁ t₁² − ½ a₂ t₂²
= ½ a₁ t₁² − ½ a₂ (−(a₁ t₁)/a₂)²
= ½ a₁ t₁² − ½ (a₁²/a₂) t₁²
= ½ ( a₁ − (a₁²/a₂)) t₁²

t₁ = √(2 d /( a₁ − (a₁²/a₂)) )

T = (1 − a₁ /a₂) √(2 d /( a₁ − (a₁²/a₂)) )

so (keeping track of the minus signs)
T = ( 1 + 6.9/5.9)√(600 / ( 6.9 + 6.9 * 6.9 / 5.9 )) = 13.7 s
with maximum velocity
v = a₁ t₁ = a₁ √(2 d /( a₁ − (a₁²/a₂)) ) = 6.9 √(600 / ( 6.9 + 6.9 * 6.9 / 5.9 )) = 43.7 m/s

Double check
43.7 / 6.9 = 6.33
43.7 / 5.9 = 7.41
6.33 + 7.41 = 13.7

13.7 * 43.7 /2 = 300

69%---- 300----- 161.71875----- D for decel----7.404047221
59%---- 300----- 138.28125-- D for accel ----6.330996899
128%------------- 300------------------13.73504

confirmed

Thanks rpenner, great help!

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