Chapter 2: Q. 42 (page 61)

A particle’s velocity is given by the function vx = 12.0 m/s2sin1pt2,where t is in s. a. What is the first time after t = 0 s when the particle reaches a turning point? b. What is the particle’s acceleration at that time?

Expert verified

Part (a)

The first time after $t = 0 s$ when a particle reaches a turning point is $t = 1 s$ .

Part (b)

Acceleration of the particle at the turning point islocalid="1649476593269" $6.3 \frac{m}{s^{2}}$ .

Step by step solution

The velocity of the particle is given by a function $v_{x} = (2.0 \frac{m}{s}) sinπ t$ .

The turning point is the point in motion where a particle reverses its direction.

It is a point where velocity is instantaneously zero and position is maximum or minimum.

The velocity is zero when

$0 = 2 sinπ t sinπ t = 0 t \Rightarrow 0 s, 1 s, . . .$

Therefore, the particle reaches its turning point after the first time at $t = 1 s$ .

Acceleration is given by $a_{x} = \frac{d v_{x}}{d t}$ .

Differentiate given expression with respect to $t$ .

localid="1649476489349" $a_{x} = \frac{d}{d t} (2 sinπ t) = 2 πcosπ t$

The acceleration of the particle at $t = 1 s$ is

localid="1649476559659" $a_{x} (a t t = 1 s) = 2 πcosπ (1) = 6.3 \frac{m}{s^{2}}$

Therefore, the acceleration of the particle at turning point islocalid="1649476571694" $6.3 \frac{m}{s^{2}}$ .

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