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Chapter 2: Q20P (page 33)

(a) If the position of a particle is given byx=20t-5t2, where x is in m and t in secwhen, if ever, the particle鈥檚 velocity is zero? (b) When is its acceleration a zero? (c) For what time range (positive or negative) is a negative? (d) Positive? (e) Graph x(t), v(t) and a(t).

Short Answer

Expert verified

(a) Particle鈥檚 velocity is zero att=1.2s

(b) Particle鈥檚 acceleration is zero att=0s

(c) Range is negative att>0

(d) Range is positive att<0

(e) Graph of xt,vt andat

Step by step solution

01

Given information

x(t)=20t-5t3

02

To understand the concept of simple mathematical operation

This problem involves simple mathematical operation of derivative. Here differentiating the given equation, the equation for velocity and acceleration can be found. Further, by pluggingand, time whereandare zero can be calculated.

Formula:

vt=dxtdt

03

Derive the expressions for velocity and acceleration

xt=20t-5t2

Expressions for velocity and acceleration can be found by taking derivative of above equation with respect to t,

vt=dxtdt=d20t-5t2dt=20-15t2at=dvtdt=d20-15t2dt=-30t

04

(a) Calculate the time at which velocity is zero

v(t)=20-15t20=20-15t2

t=2015=1.2s

05

(b) Calculate the time at which acceleration is zero

at=-30t0=-30tt=0s

06

(c)Calculate the time range in which acceleration is negative

From equation, if acceleration a(t)=-30tatt>0 acceleration is negative.

07

(d) Calculate the time range in which acceleration is positive

From equation, if accelerationa(t)=-30t att<0 acceleration is positive.

08

(e) Graph x(t), v(t) and a(t)

Graphs are shown below:

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