Math Problem Statement

Question 2: (a) (i) In a television tube, an electron travels in a straight line from the electron gun to the fluorescence screen, a distance of 46 cm, in 6 nanoseconds. What is the uniform speed? (ii) A golf ball is released from rest from the top of a very tall building. Neglecting air resistance, calculate the position and velocity of the ball after 1 s. (b) A velocity function describing the motion of a car is given by v = m + nt^2, where m = 10 m/s and n = 2 m/s^3. (i) Find the change in velocity of the car for the interval between t1 = 2 s and t2 = 5 s. (ii) Find the average acceleration in this time interval. (iii) Find the instantaneous acceleration at time t = 2 s. (c) A long jumper leaves the ground at an angle of 20° to the horizontal and at a speed of 11 m/s. Regard the jumper as a particle. How far does the jumper jump?

Solution

Let’s solve the questions one by one.


(a) (i)

  • Given:
    • Distance = 46cm=0.46m46 \, \text{cm} = 0.46 \, \text{m}
    • Time = 6ns=6×109s6 \, \text{ns} = 6 \times 10^{-9} \, \text{s}
  • To Find: Uniform speed vv
  • Solution: v=DistanceTimev = \frac{\text{Distance}}{\text{Time}} v=0.466×109=7.67×107m/sv = \frac{0.46}{6 \times 10^{-9}} = 7.67 \times 10^7 \, \text{m/s}

(a) (ii)

  • Given:
    • The golf ball is released from rest (u=0u = 0).
    • Acceleration due to gravity g=9.8m/s2g = 9.8 \, \text{m/s}^2.
    • Time t=1st = 1 \, \text{s}.
  • To Find:
    1. Position ss after 1s1 \, \text{s}.
    2. Velocity vv after 1s1 \, \text{s}.
  • Solution: Using equations of motion: s=ut+12gt2s = ut + \frac{1}{2} g t^2 s=0+12(9.8)(12)=4.9ms = 0 + \frac{1}{2} (9.8) (1^2) = 4.9 \, \text{m} v=u+gtv = u + g t v=0+9.8(1)=9.8m/sv = 0 + 9.8 (1) = 9.8 \, \text{m/s}

(b)

Given: v=m+nt2,m=10m/s,n=2m/s3v = m + nt^2, \quad m = 10 \, \text{m/s}, \, n = 2 \, \text{m/s}^3

(i) Change in velocity

  • At t1=2st_1 = 2 \, \text{s}: v1=10+2(22)=10+8=18m/sv_1 = 10 + 2(2^2) = 10 + 8 = 18 \, \text{m/s}
  • At t2=5st_2 = 5 \, \text{s}: v2=10+2(52)=10+50=60m/sv_2 = 10 + 2(5^2) = 10 + 50 = 60 \, \text{m/s}
  • Change in velocity: Δv=v2v1=6018=42m/s\Delta v = v_2 - v_1 = 60 - 18 = 42 \, \text{m/s}

(ii) Average acceleration

  • Formula: Average acceleration=ΔvΔt\text{Average acceleration} = \frac{\Delta v}{\Delta t}
  • Solution: aavg=4252=423=14m/s2a_{\text{avg}} = \frac{42}{5 - 2} = \frac{42}{3} = 14 \, \text{m/s}^2

(iii) Instantaneous acceleration at t=2st = 2 \, \text{s}

  • Acceleration is the derivative of velocity: a=ddt(m+nt2)a = \frac{d}{dt} (m + nt^2) a=0+2nta = 0 + 2n t
  • At t=2st = 2 \, \text{s}: a=2(2)(2)=8m/s2a = 2(2)(2) = 8 \, \text{m/s}^2

(c)

  • Given:
    • Initial velocity u=11m/su = 11 \, \text{m/s}
    • Angle of projection θ=20\theta = 20^\circ
    • Acceleration due to gravity g=9.8m/s2g = 9.8 \, \text{m/s}^2
  • To Find: Horizontal range (RR).
  • Solution: Horizontal range is given by: R=u2sin2θgR = \frac{u^2 \sin 2\theta}{g} sin2θ=sin(2×20)=sin40=0.6428\sin 2\theta = \sin(2 \times 20^\circ) = \sin 40^\circ = 0.6428 R=(11)2(0.6428)9.8R = \frac{(11)^2 (0.6428)}{9.8} R=121(0.6428)9.8=77.789.8=7.94mR = \frac{121 (0.6428)}{9.8} = \frac{77.78}{9.8} = 7.94 \, \text{m}

Would you like further details or clarification?

Related Questions

  1. What factors affect the velocity of a golf ball in free fall?
  2. How does the angle of projection impact the range of a projectile?
  3. What is the significance of instantaneous acceleration in motion problems?
  4. Can we find the time of flight for the long jumper? How?
  5. How would the motion change if air resistance is considered?

Tip: Always break down problems into known formulas and steps to minimize confusion.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Projectile Motion
Derivatives
Velocity and Acceleration

Formulas

Uniform speed: v = distance/time
Equations of motion: s = ut + (1/2)gt^2 and v = u + gt
Change in velocity: Δv = v2 - v1
Average acceleration: a_avg = Δv/Δt
Instantaneous acceleration: a = d(v)/dt
Horizontal range for projectile motion: R = (u^2 * sin(2θ)) / g

Theorems

Basic Kinematic Equations
Derivative for instantaneous rate of change

Suitable Grade Level

Grades 11-12