Figure 1 of 1The plot shows the position x of an object as a function of time t.

Question

Figure

1 of 1The plot shows the position x of an object as a function of time t. Time varies from 0 to 17 seconds on the horizontal axis. The position is from 0 to 14 meters on the vertical axis. The position is shown to increase linearly from the origin to 3 seconds and 7 meters. It stays at this value till 6 seconds. Then, it increases linearly to 11 seconds and 10 meters and then, decreases linearly to 16 seconds and 7 meters. Finally, the position decreases linearly to 17 seconds and 0 meters.

a) Find the object's positions x1, x2, x3, and x4 at times t1=2.0s, t2=4.0s, t3=13s, and t4=17s.

b) Find the object's speeds v1, v2, and v3 at times t1=2.0s, t2=4.0s, and t3=13s.

At any moment in time, the object can be stationary, moving to the right, moving to the left, or turning around. Identify the motion of the object at each time provided.

Explanation / Answer

a)

Slope of the curve between t = 0 to t = 3s,

m = 7/3 = 2.33 m/s

So, x1 = 2.33*2 = 4.67 m <------- answer

x2 = 7 m <-------- answer

x3 = 10 + ((7 - 10)/(16 - 11))*2

= 8.8 m <--------- answer

b)

v1 = slope at t = 2s

So, v1 = 7/3 = 2.33 m/s <------ answer

v2 = slope at t = 4s

So, v2 = 0

Similalrly, v3 = ((7 - 10)/(16 - 11)) = -0.6 m/s

c)

The particle moves to right between t = 0 to t = 3s as the slope is positive

The particle then remains stationary between t =3s to t = 6s

Then again the particle moves towards right between t = 6s to t = 11s

The particle then moves towards left from t =11 s to t = 17 s as the slope of the curve is negative

So, the turn around point is t = 11 s

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