Chapter 2: Q. 37 (page 61)

Particles A, B, and C move along the x-axis. Particle C has an initial velocity of 10 m/s. In FIGURE P2.37, the graph for A is a position-versus-time graph; the graph for B is a velocity-versus time graph; the graph for C is an acceleration-versus-time graph. Find each particle’s velocity at t = 7.0 s.

Short Answer

Expert verified

The velocity of particle A at $t = 7 s$ is $- 10 \frac{m}{s}$ .

The velocity of particle B at $t = 7 s$ is $- 20 \frac{m}{s}$ .

The velocity of particle c at $t = 7 s$ is $95 \frac{m}{s}$ .

Step by step solution

Particle A

The position-versus-time graph is given for particle A.

The velocity is given by the slope of the position-versus-time graph.

The graph for particle A is a straight line from $t = 5 s$ to $t = 8 s$ .

Consider points $(t_{1}, x_{1}) = (5, 0)$ and $(t_{2}, x_{2}) = (8, - 30)$

The velocity of particle A at $t = 7 s$ is

$v = \frac{x_{2} - x_{1}}{t_{2} - t_{1}} = \frac{- 30 - 0}{8 - 5} = \frac{- 30}{3} = - 10 \frac{m}{s}$

Therefore, the velocity of particle A at $t = 7 s$ is $- 10 \frac{m}{s}$ .

Particle B

The velocity-versus-time graph is given for particle B.

From the graph of a velocity-versus-time, it can be observed that the velocity of particle B is $- 20 \frac{m}{s}$ .

Particle C

The acceleration-versus-time graph is given for particle C.

The velocity of a particle $∆ v = v_{f} - v_{0}$ between $t_{0}$ and $t_{f}$ is the area under the curve from $t_{0}$ to $t_{f}$ .

Thus, the area under the acceleration - time curve is

$∆ v =$ area of the rectangle between $t = 0 s$ and role="math" localid="1648480759533" $t = 2 s +$ area of the triangle between $t = 2 s$ and role="math" localid="1648480812177" $t = 5 s -$ area of the triangle between $t = 5 s$ and $t = 7 s$

$∆ v = (2 s) (30 \frac{m}{s^{2}}) + \frac{1}{2} (3 s) (30 \frac{m}{s^{2}}) - \frac{1}{2} (2 s) (20 \frac{m}{s^{2}}) = 60 + 45 - 20 = 60 + 25 = 85 \frac{m}{s}$

Now,

The velocity of a particle at $t = 7 s$ is

$v_{f} = v_{0} +$ area under the acceleration-time curve

$v_{f} = 10 \frac{m}{s} + 85 \frac{m}{s} = 95 \frac{m}{s}$

Therefore, the velocity of particle C is $95 \frac{m}{s}$ .

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Particle A

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