The front of an automobile moving with speed 44 ft/s (about 30mph) is 160 ft fro

Question

The front of an automobile moving with speed 44 ft/s (about 30mph) is 160 ft from an intersection that is 65 ft wide when a green light at the intersection changes to yellow. The light will stay yellow for 4 sec. If the driver floors the accelerator it will impart an acceleration of 5.0 ft/s^2. If the driver brakes it wil impart a deceleration of 7.5 ft/s^2. The automobile is 11 ft long. a. Find the position of the automobile relative to the intersection if he floors it. Will the entire care make it all the way through the intersection before the light turns red? b. Find the position of the automobile relative to the intersection if he stands on the brake. Can he stop short of the intersection if he hits the brakes?

Explanation / Answer

The automobile (front of it, to be clear) is at a distance of 160ft from the intersection

speed of automobile = 44ft/sec

a) when the driver floors the accelerator, it will impart a constant acceleration of 5 ft/s^2.

The automobile has got 4 sec, before the signal turns red. Lets calculate how much distance it could travel with an initial speed of 44 ft/s and acceleration of 5ft/s^2 in 4 secs

from the kinematic equation

s = u*t + 1/2 * a*t^2

distance travelled = 44*4 + 1/2*5*16 = 176 + 40 = 216ft

Start of the intersection was at 160 ft from the car. Hence, by the time signal turns red, the car has travelled (216 - 160)ft = 56 ft into the intersection.

The intersection is 65 ft wide, so the car will be in the middle of intersection when signal turns red.

b)

when the driver brakes, it will impart deceleration of 7.5ft/s^2

Calculate distance travelled by the car, before coming to a halt (final velocity = 0)

from the kinematic relation v2 - u2 = 2*a*s

-(442) = 2*(-7.5)*S => S = 129.06 ft

Therefore, the car stops (160 - 129.06) ft = 30.94 ft before the intersection

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