Physic kinematic problem, please help!?

Speed = 15 * t – 2 * t^2

The truck will reach its maximum speed, when its acceleration is 0 m/s^2. The equation for acceleration versus time is the first derivative of this equation.

a = 15 – 4 * t
0 = 15 – 4 * t
t = 3.75 seconds

This is time when the truck reaches its maximum speed. To determine the truck’s maximum speed, use this time in the first equation.

Speed = 15 * 3.75 – 2 * 3.75^2 = 28.125 m/s

For the first 3.75 seconds, the equation for distance versus time is the integral of the equation of velocity versus time.

d = 7.5 * t^2 – ⅔ * t^3
Use 3.75 seconds for the time.
d = 7.5 * 3.75^2 – ⅔ * 3.75^3 = 70.3215 meters

Four minutes is 240 seconds. During this time, the truck’s speed is 28.125 m/s.

d = 28.125 * 240 = 6,750 meters

Use the following equation to determine the distance the truck moves as its velocity decreases from 28.125 m/s to 0 m/s at the rate of 2 m/s each second.

vf^2 = vi^2 + 2 * a * d
0 = 28.125^2 + 2 * -2 * d
d = 791.015625 ÷ 4 = 197.7539063 meters

Total distance = 70.3215 + 6,750 + 197.7539063 = 7,018.075406
This rounds to 7,018 meters. I hope this is helpful for you.

(a)
Method 1 : Completing square

v(t) = 15t - 2t²
v(t) = -2(t² - 7.5t)
v(t) = -2(t² - 7.5t + 3.75²) + 2*3.75²
v(t) = -2(t - 3.75)² + 28.125

For all real values of t, -2(t - 3.75)² ≤ 0
Hence, v(t) = -2(t - 3.75)² + 28.125
When t = 3.75, maximum v(t) = 28.125 (m/s)

Method 2 : Differentiation

v(t) = 15t - 2t²
v'(t) = 15 - 4t = -4(t - 3.75)
v"(t) = -4

When t = 3.75 : v'(t) = 0 and v"(t) = -4 < 0
Hence, maximum v(t) at t = 3.75
Maximum v(t) = 15(3.75) - 2(3.75)² = 28.125 m/s ≈ 28.1 m/s

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(b)
Refer to part (a).
Maximum v(t) at t = 3.75 s

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(c)
Distance traveled on accelerating
= ∫[0,3.75] (15t - 2t²)dt
= [(15/2)(3.75)² - (2/3)(3.75)³] - [(15/2)(0)² - (2/3)(0)³]
= 70.3125 m
≈ 70 m

Distance traveled on constant speed for 4 minutes
= 28.125 * 4 * 60
= 6750 m

v² = vₒ² + 2as
s = (v² - vₒ²)/(2a)
Distance traveled on decelerating, s
= (0 - 28.125)/[2*(-2)]
= 7.03125 m
≈ 7 m

Distance traveled during the entire trip
= 70 + 6750 + 7
= 6827 m

v = A t - B t^2
v is max when a = dv/dt = 0
then
A - 2B t = 0
t = A / 2B = 15/4 = 3.75 s [b]
max speed is
v' = 15*3.75 - 2*3.75^2 = 28.125 m/s [a]
space traveled during entire trip is
s = s1 + s2 + s3
s1 = 1/2 A t1^2 - 2/3 B t1^3 = 1/2*15*3.75^2 - 2/3*2*3.75^3 = 35.15625 m
s2 = v' t2 = 28.125*(4*60) = 6750 m
s3 = v'^2 / (2 a) = 28.125^2 / (2*2) = 197.75391
then
s = 6982.91016 m [c]

Hi