1) The Distance between school and a student\'s home is 15 miles. During a typic

Question

1) The Distance between school and a student's home is 15 miles. During a typical school day it takes 45 minutes for this student to drive from home to school in the morning and 30 minutes from school to home in the afternoon.

a) What is the average speed of this student during a school day (in m/h)?

2) The velocity of a particle is given as v(t) = 1.0t^2 - 3.0t - 4.0 m/s , where t is in second and t > or = 0.

a) When the particle will be temporary at rest?

3) The position of an object moving along a straight line is given by x(t)= -3.0t^3 + 2.0t^2 - 1.0t, where x is in meters, t > or equal 0, t is in seconds. Will this object be momentarily at rest?  ............   Why?

4) Step 1: Solve x = va + at for t

     Step 2: Substitute t into x - x0 = v0t + 1/2at^2

     Step 3: Sort out the expression so that you can get v^2 = v0^2 + 2a(x-x0)

Explanation / Answer

1) total to and fro distance = 15+15 = 30 miles

total time taken = 45+30 = 75mis = 1.25hts

so average speed = 30/1.25 = 24 miles/hr

2)v(t) = 0 for particle to be at rest

so 1.0t^2 - 3.0t - 4.0 m/s = 0

so t = 4secs ( as t is only positive)

so at 4 secs, particle will be at rest tempororily

3) velocity = dx/dt

x(t)= -3.0t^3 + 2.0t^2 - 1.0t

so v(t) = -9t^2 + 4t -1v(t) = 0 for particle to be at rest

so -9t^2 + 4t -1 m/s = 0so t = in imaginory

so not possible, particle will not be at rest at any time tempororily

4) t = (x-va)/a

x- x0 = v0(x-va)/a + (1/2)(a)(x-va)^2/a^2

so 2a(x - x0) = 2v0(x-va) + (x-va)^2

so 2a(x-x0) = (x-va)(2v0 + x - va)

so 2a(x-x0) = (v + v0)(v-v0)

so 2a(x-x0) = v^2 - v0^2

so v^2 = v0^2 + 2a(x-x0)

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