Calculus I - Antiderivative Test Question. help!!!?

Question

the question is:::

A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 38 ft/s2. What is the distance covered before the car comes to a stop?

please explain w/ answer please!! thanky you

Answer 1

Because the speed is given in miles per hour, first you need to get it in ft per second. Now,

60 m.p.h. = 88 f.p.s. so

10 m.p.h. = 14.67 f.p.s. so

50 m.p.h. = 14.67 x 5 = 73.3 f.p.s.

Now use v^2 = u^2 - 2as, where v is the final velocity, zero in this case, u the initial velocity, 73.3, a the deceleration, 38 ft/sec^2, and s the distance, so

v^2 = u^2 - 2as,

0 = (73.3)^2 - 2 x 38 x s, so

5378 = 76s, so

s = 5378/76, so

s = 70.67 feet.

Hope this helps, Twiggy.

Answer 2

Hello,
Second derivative of distance with respect to time is acceleration. So d2s/d2t = -38 Then the antiderivative is ds/dt = -38t + c which gives us the velocity. Then the velocity at time 0 is 50 mph or 73 13 ft/s so we have c = 73 1/3. Now ds/dt = -38t + 731/3. Then the antiderivitave is s = -19t^2 + 73 1/3 t + c this is the distance function. At time = the distance = 0 so c = 0.

Now the velocity = 0 when the car stops so in the second derivative we have 0 = -38t + 73 1/3 so t = (73 1/3)/38 = 1.93 then we put this in for t in the distance equation and we have

s = -19*(1.93)^2 + 50*(1.93) = 25.7 ft.

Hope This Helps!

Answer 3

since the acceleration is constant, the velocity v at time t
v = 50 -38t
set v= 0 to find when the car stops.
t = 50/38 (so after about 4/3 seconds).
now the distance s is the antiderivative of v (v = ds/dt)
so s = 50t - 19t^2 (easy check that v = ds/dt)
plug in t = 50/38
s = 2500/38 -19(2500/38^2) = 2500/38 - 2500/76 = 2500/38 - 1250/38 = 1250/38 =625/19 feet, about 32.9