Differential Equations?

Question

Help me to solve these three problems:

1.If a car goes from 0 to 80mph in six seconds with constant
acceleration, what is that acceleration?
2.An Acura NSX going at 70mph stops in 157 feet. Find the
acceleration, assuming it is constant.
3.A car going 80ft/sec (about 55 mph) brakes to stop in five
seconds. Assume the deceleration is constant. Find the total
distance traveled using antidifferentiation.

Answer 1

1.
Convert 80 mph to feet per second: 80 * 5280 / 3600 = 117.333

a = (Vf - Vo) / t = (117.333 - 0) / 6 = 19.556 ft /sec^2

2.

Convert 70 mph to feet per second: 70 * 5280 / 3600 = 102.667

Vf^2 = V0^2 - 2*a*d

0 = 102.667^2 - 2*a*157

10540.44 = 314 a

a = 33.57 ft / sec^2


3. a = (Vf - Vo) / t = 80 / 5 = 16 ft/sec^2

V = 80 - 16t

Integrate that equation from 0 to 5

80t - 8t^2 = 80 * 5 - 8*25 = 200 ft

Answer 2

1.
80 mph = 35.7632 meters / second
acceleration = change of velocity / time = 35.7632 meters / second / 6seconds = 5.96 m/s2.

2.
31.2928 meters / second
157 feet = 47.8536 meters

x = a/2 * t*t
a = v/t

so x = v*v/2/a
a = v*v/2/x = 10.23 m/s2.

3.
80 feet / sec = 24.38400 meters / sec
again:
x = a/2 * t*t
a = v/t
so
x=v*t/2 = 60.96 meters.