Question #76235

An object starts from rest and accelerates for 0.24 meter per second square for 2 minutes and continues at a steady speed for 3 minutes and slow down to stop for 1 minute. a. Draw graph for the motion. b. Calculate the maximum velocity in km/hr. c. Determine the retardation. d. find the displacement in 5 minutes.

Expert's answer

Answer on Question #76235 Physics / Atomic and Nuclear Physics

An object starts from rest and accelerates for a=0.24a = 0.24 meter per second square for t1=2t_1 = 2 minutes and continues at a steady speed for t2=3t_2 = 3 minutes and slow down to stop for t3=1t_3 = 1 minute. A) Draw graph for the motion. B) Calculate the maximum velocity in km/hr. C) Determine the retardation. D) Find the displacement in 5 minutes.

Solution:


A)

B) vmax=at1=0.24×2×60=28.8ms=103.68kmhv_{\mathrm{max}} = at_1 = 0.24 \times 2 \times 60 = 28.8 \frac{\mathrm{m}}{\mathrm{s}} = 103.68 \frac{\mathrm{km}}{\mathrm{h}}

C) a3=0vmaxt3=028.860=0.48ms2a_3 = \frac{0 - v_{\mathrm{max}}}{t_3} = \frac{0 - 28.8}{60} = -0.48 \frac{\mathrm{m}}{\mathrm{s}^2}

D) s=vmax2t1+vmaxt2=28.82×2×60+28.8×3×60=6912ms = \frac{v_{\mathrm{max}}}{2} t_1 + v_{\mathrm{max}}t_2 = \frac{28.8}{2}\times 2\times 60 + 28.8\times 3\times 60 = 6912\mathrm{m}

Answers: 103.68kmh103.68 \frac{\mathrm{km}}{\mathrm{h}} , 0.48ms2-0.48 \frac{\mathrm{m}}{\mathrm{s}^2} , 6912 m6912 \mathrm{~m}

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