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Assoli18 [71]
3 years ago
7

Ms. Terry Novel is driving down the highway at 26 m/s when the driver sees the light ahead turn yellow. It takes the driver 0.3

s to react and the maximum braking acceleration of her car is 5.3 m/s2. If the light stays yellow for a total of 4 seconds, how far does she travel in this time?
Physics
1 answer:
Ket [755]3 years ago
7 0

Answer:

67.8 m.

Explanation:

For 0.3 sec which is the response time , he will keep on moving with velocity

undiminished or at 26 m/s .

Distance travelled during this period

= 26 x .3 =7.8 m

For rest of the time ie during 4 - .3 = 3.7 s , he decelerates with deceleration of 5.3 ms⁻²  and initial velocity of 26 m/s

s = u t - 1/2 a t²

s = 26 x 3.7 - .5 x 5.3 x 3.7 x 3.7

= 96.2 - 36.28

= 60 m approx.

Total distance travelled

=60 + 7.8 = 67.8 m.

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Read 2 more answers
A dolphin in an aquatic show jumps straight up out of the water at a velocity of 15.0 m/s. (a) List the knowns in this problem.
astra-53 [7]

Answer:

a)

Y0 = 0 m

Vy0 = 15 m/s

ay = -9.81 m/s^2

b) 7.71 m

c) 3.06 s

Explanation:

The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards

Y(0) = 0 m

Vy(0) = 15 m/s

ay = -9.81 m/s^2 (negative because it points down)

Since acceleration is constant we can use the equation for uniformly accelerated movement:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

To find the highest point we do the first time derivative (this is the speed:

V(t) = Vy0 + a * t

We equate this to zero

0 = Vy0 + a * t

0 = 15 - 9.81 * t

15 = 9.81 * t

t = 0.654 s

At this time it will have a height of:

Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m

The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.

0 = Y0 + Vy0 * t + 1/2 * a * t^2

0 = 0 + 15 * t - 1/2 * 9.81 t^2

0 = 15 * t - 4.9 * t^2

0 = t * (15 - 4.9 * t)

t1 = 0 This is the moment it jumped into the air

0 = 15 - 4.9 * t2

15 = 4.9 * t2

t2 = 3.06 s This is the moment when it falls again.

3.06 - 0 = 3.06 s

5 0
3 years ago
Please don’t troll. I need someone to actually answer these questions.
mash [69]

Answer:

-3+3 i think this is the answer

Explanation:

i think you can ask someone else sorry

4 0
2 years ago
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