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drek231 [11]
3 years ago
15

4. You are walking at a leisurely pace of .75 m/s when you hear the train coming and start running to catch it. Over the next 10

seconds, you speed up to 3.5 m/s and catch your train! What was your acceleration over that time?
Physics
1 answer:
hjlf3 years ago
5 0

Answer:

0.275 m/s²

Explanation:

The acceleration of an object is defined as the change in velocity over the change in time.

  • a = Δv/Δt
  • a = (v - v₀)/t

We are given the initial velocity, 0.75 m/s. The final velocity of the user in question is 3.5 m/s.

The time it takes for us to catch the train is 10 seconds. Using these variables, let's substitute into our equation.

  • a = (3.5 - 0.75)/10
  • a = 2.75/10
  • a = 0.275 m/s²

Our acceleration over that time was 0.275 m/s² in order to catch the train.

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An automobile with a standard differential turns sharply to the left. The left driving wheel turns on a 20-m radius. Distance be
Inessa05 [86]

Explanation:

The given data is as follows.

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   Distance between left and right wheel = 1.5m,

Let us assume speed of drive shaft is N rpm.

Formula to calculate angular velocity is as follows.

    Angular velocity of automobile = w = \frac{V}{R}

where,   V = linear velocity of automobile m/min,

              R = turning radius from automobile center in meter

In the given case, angular velocity remains same for inner and outer wheel but there is change in linear velocity of inner wheel and outer wheel.

Now, we assume that

         u = linear velocity of inner wheel

and,   u' = linear velocity of outer wheel.

Formula for angular velocity of inner wheel w = ,

Formula for angular velocity of outer wheel w =

Now, for inner wheels

                   w =

                      = \frac{u}{(R - d)}

                  u = V \times \frac{(R - d)}{R}

                    = V \times (1 - \frac{d}{R})

If radius of wheel is r it will cover  distance in one min.

Since, velocity of wheel is u it will cover distance u in unit time(min)

Thus,             u = 2\pi rn = V \times (1 - \frac{d}{R})

Now, rotation per minute of inner wheel is calculated as follows.

         n = \frac{V}{2 \pi r \times (1 - \frac{d}{R})}

            = \frac{V}{2 \pi r \times (1 - \frac{0.75}{20})} (since 2d = 1.5m given, d = 0.75m),

             = \frac{V}{r} \times 0.1532

So, rotation per minute of outer wheel; n' =  

                   = \frac{V}{2 \pi r \times (1 + \frac{0.75}{20})}

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5 0
4 years ago
A 200kg ball on the end of string is swung in horizontal circle with radius of 0.5m . The ball makes revolution every 2second th
ANEK [815]

Answer:\dfrac{\pi}{2} ms^{-1}

Explanation:

Let T be the time required to make one revolution.

Let r be the radius of the circular path.

Let d be the distance travelled by ball in one revolution.

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Speed of an object moving is circular path is define as the ratio of distance travelled in one revolution to the time taken by the object to complete one revolution.

Let s be the speed of the ball.

s=\frac{d}{T}=\frac{\pi }{2}ms^{-1}

So,the speed of the ball is \frac{\pi }{2}ms^{-1}

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