Question

Q1. A truck traveling at 40 mph is approaching a stop sign. At time ????0 and at a distance of 80ft, the truck begins to slow down by decelerating rate of 12 ft/sec2 . Will the truck be able to stop in time?

Answer 1

Answer:

The truck will not stop in time. The truck passes the stop sign by about 63.41 ft before it stops.

Explanation:

The distance that the truck starts slowing down = 80 ft from the stop sign

Using equations of motion, we can calculate the distance it will take the truck to stop, then check of it is less than or more than 80 ft.

u = initial velocity of the truck = 40 mph = 58.667 ft/s

v = final velocity of the truck = 0 ft/s (since it comes to a stop eventually)

x = horizontal distance covered during the deceleration

a = Deceleration = -12 ft/s² (it'll have a negative sign, since it is negative acceleration

v² = u² + 2ax

0² = 58.667² + 2(-12)(x)

24x = 3441.816889

x = 143.41 ft

143.41 ft > 80 ft; hence, the truck will not stop in time. The truck passes the stop sign by about 63.41 ft before it stops.

Answer 2

Corrected Question:

A truck traveling at 40 mph is approaching a stop sign. At time t₀ and at a distance of 80 ft, the truck begins to slow down by decelerating at 12 ft/s2, will the truck be able to stop in time?

Answer:

The truck will not be able to stop in time.

Explanation:

==> First lets convert all variables to SI units

1 mph = 0.45m/s

40mph = 40 miles per hour = 40 x 0.45 m/s

40mph = 18m/s

1 ft = 0.3048m

80 ft = 80 x 0.3048m

80 ft = 24.38m

Also;

12ft/s² = 12 x 0.3048m/s²

12ft/s² = 3.66m/s²

==> Now, consider one of the equations of motion as follows;

v² = u² + 2as               -----------------(i)

Where;

v = final velocity of motion

u = initial velocity of motion

a= acceleration/deceleration of motion

s = distance covered during motion

Using this equation, lets calculate the distance, s, covered during the acceleration;

We know that;

v = 0               [since the truck comes to a stop]

u = 40mph = 18m/s

a = -12ft/s² = -3.66m/s²    [the negative sign shows that the truck decelerates]

Substitute these values into equation (i) as follows;

0² = 18² + 2 (-3.66)s

0 = 324 - 7.32s

7.32s = 324

s = $\frac{324}{7.32}$

s = 44.26m

The distance from where the truck starts decelerating to where it eventually stops is 44.26m which is past the stop sign (which is at 80ft = 24.38m).  This means that the truck stops, 44.26m - 24.38m = 19.88m, after the stop sign. Therefore, the truck will not be able to stop in time.