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tester [92]
3 years ago
6

A motorist traveling with a constant speed of 15 m/s (about 34 mi/h) passes a school-crossing corner, where the speed limit is 0

m/s (about 22 mi/h). Just as the motorist passes, a police officer on a motorcycle stopped at the comer starts off in pursuit with constant acceleration of 3.0 m/S2. (a) How much time elapses before the officer catches up with the motorist? (b) What is the officer's speed at that point? (c) What is the total distance each vehicle has traveled at that point? Please help me
Physics
1 answer:
nikklg [1K]3 years ago
6 0

Answer:

(a) 10 s

(b) 30 m/s

(c) 150 m

Explanation:

The motorist's position at time t is:

x = 15t

The officer's position at time t is:

x = ½ (3) t² = 1.5 t²

(a) When they have the same position, the time is:

15t = 1.5 t²

t = 0 or 10 s

(b) The officer's speed is:

v = 3t

v = 30 m/s

(c) The position is:

x = 15t = 150 m

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DiKsa [7]

Answer:(a)360N,(b)171N,(c)2.702m

Explanation:

(a)Maximum Friction Force =\mu \left ( N\right )=0.4\times \left ( 740+160\right )

=360 N

cos\theta =\frac{3}{5}

sin\theta =\frac{4}{5}

(b)Moment about Ground Point

740\times 1\times cos\theta +2.5\times 160\times cos\theta -N_15sin\theta

N_1tan\theta =1140

N_1=171 N

N_1=f=171 N

(c)

740\times x\times cos\theta +2.5\times 160\times cos\theta -N_15sin\theta

Here maximum friction force can be 360 N

Therefore N_1=360 N

Where x is the maximum distance moved by man along the ladder

360\times 5\times \frac{4}{3}=740x+160\times 2.5

740x=2000

x=2.702m

5 0
3 years ago
The velocity of a particle moving along the x-axis varies with time according to v(t) = A + Bt−1 , where A = 2 m/s, B = 0.25 m,
kondaur [170]

Answer:

a= -2\ m/s^2

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v=2+0.25t^{-1}

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we know that

a=\dfrac{dv}{dt}

\dfrac{dv}{dt}=0-0.5t^{-2}

a=-0.5t^{-2}

Acceleration at t= 2 s

a=-0.5\times 2^{-2}

a= -2\ m/s^2

Acceleration at t= 5 s

a=-0.5\times 5^{-2}

a=-12.5\ m/s^2

We know that

v=\dfrac{dx}{dt}

dx=\left(2+\dfrac{1}{4t}\right)dt

Position at t= 2 s:

\int_{0}^{x}dx=\int_{1}^{2} \left(2+0.25\dfrac{1}{t}\right)dt

x=\left [2t+0.25\ lnt \right ]_{1}^{2}

x=2+0.25 ln2

x=2.17 m

Position at t= 5 s:

\int_{0}^{x}dx=\int_{1}^{5} \left(2+0.25\dfrac{1}{t}\right)dt

x=\left [2t+0.25\ lnt \right ]_{1}^{5}

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inysia [295]
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When a 12 V battery is connected to a 6 uF capacitor, how much energy is stored
adoni [48]

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E = (1/2) · (6 x 10⁻⁶ F) · (12 V)²

E = (3 x 10⁻⁶ F) · (144 V²)

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An electron starts from rest in a vacuum, in a region of strong electric field. The electron moves through a potential differenc
Semenov [28]

Answer:

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The  negative sign is not includes because kinetic energy can not be negative

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