A motorist is driving at 20m/s when she sees that a traffic light 200m ahead has just turned red. She knows that this light stays red for 15s, and she wants to reach the light just as it turns green again. It takes her 1.0s to step on the brakes and begin slowing. What is her speed as she reaches the light at the instant it turns green?

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 200-20=20\times 14+\frac{1}{2}\times a\times 14^2\\\Rightarrow a=\frac{2(180-20\times14)}{14^2}\\\Rightarrow a=-1.02040816327\ m/s^2[/tex]

[tex]v=u+at\\\Rightarrow v=20-1.02040816327\times 14\\\Rightarrow v=5.71428571422\ m/s[/tex]

The speed as she reaches the light at the instant it turns green is 5.71428571422 m/s

asuwafohjames asuwafohjames · Answer 2 · 2022-01-27T11:58:24+01:00

The final speed of the motorist as she reaches the light is 5.72 m/s

To calculate the speed of the motorist as she reaches the light, we need to first find the deceleration of the motorist.

Formula:

S₁ = ut₁.............. Equation 1

Where:

S₁ = displacement of the motorist as its slows down
u = initial velocity
t₁ = time it takes to slow down.

Given:

u = 20 m/s

t₁ = 1

Therefore,

S₁ = 20(1) = 20 m

Then,

S = 200-20
S = 180 m

Where:

S = distance covered by the motorist before the brake is applied.

we use the formula below to calculate the deceleration of the motorist.

S = ut+at²/2........... Equation 2

Where:

a = deceleration of the motorist.

Given:

u = 20 m/s
t = 15-1 = 14 s
S = 180 m

Substitute these values into equation 2

180 = 20(14)+a(14²)/2
180 = 280+98a

Solve for a

98a = 180-280
98a = -100
a = -100/98

a = -1.02 m/s²

Finally, we use the formula below to get the speed of the motorist as she reaches the light.

v = u+at............ Equation 3

Given:

u = 20 m/s
a = -1.02 m/s²
t = 14 s

Substitute these values into equation 3

v = 20+(-1.02×14)
v = 20-14.28
v = 5.72 m/s.

Hence, The final speed of the motorist as she reaches the light is 5.72 m/s

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