A jogger accelerates from rest to 5.23 m/s in 3.76 s. A car accelerates from 18.

Question

A jogger accelerates from rest to 5.23 m/s in 3.76 s. A car accelerates from 18.1 to 32.3 m/s also in 3.76 s (a) find the magnitude of the acceleration of the jogger. (B) determine the magnitude of the acceleration of the car. (C) how much further does the car travel than the jogger during the 3.76 s? A jogger accelerates from rest to 5.23 m/s in 3.76 s. A car accelerates from 18.1 to 32.3 m/s also in 3.76 s (a) find the magnitude of the acceleration of the jogger. (B) determine the magnitude of the acceleration of the car. (C) how much further does the car travel than the jogger during the 3.76 s?

Explanation / Answer

now we find the magnitude of the acceleration of the jogger

acceleration of the jogger a1=v-u/t=5.23-0/3.76=1.4 m/s^2

now we find the magnitude of the acceleration of the car

acceleration of the car a2=32.3-18.1/3.76=3.78 m/s^2

now we find the distance

distance s=ut+1/2at^2

=18.1*3.76+1/2*3.78*3.76^2=94.78 m

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