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solmaris [256]
3 years ago
8

A skydiver jumps from an airplane and falls freely for 5 seconds. If the sky diver's velocity increases by 49 m/s during that ti

me, at which rate is the diver accelerating ?
Physics
2 answers:
Sergeeva-Olga [200]3 years ago
8 0

Acceleration  =  (change in speed)  /  (time for the change)

                       =      (49 m/s)   /   (5 seconds)

                       =       (49 / 5)  m/s  /  s

                       =            9.8 m/s² 
mihalych1998 [28]3 years ago
3 0
Your answer would be 9.8
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At an instant a traffic light turns green an automobile that has been waiting at an intersection of the road accelerates with 5m
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1) The car overtakes the truck at a distance of 160 m far from the intersection.

2) The velocity of the car is 40 m/s

Explanation:

1)

The car is travelling with a constant acceleration starting from rest, so its position at time t (measured taking the intersection as the origin) is given by

x_c(t) = \frac{1}{2}at^2

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t is the time

On the other hand, the truck is travelling at a constant velocity, therefore its position at time t is given by

x_t(t) = vt

where

v = 20 m/s is the velocity of the truck

t is the time

The car overtakes the truck when the two positions are the  same, so when

x_c(t) = x_t(t)\\\frac{1}{2}at^2 = vt\\t=\frac{2v}{a}=\frac{2(20)}{5}=8 s

So, after a time of 8 seconds. Therefore, the distance covered by the car during this time is

x_c(8) = \frac{1}{2}(5)(8)^2=160 m

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2)

The motion of the car is a uniformly accelerated motion, so the velocity of the car at time t is given by the suvat equation

v=u+at

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v is the velocity at time t

u is the initial velocity

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For the car in this problem, we have:

u = 0 (it starts  from rest)

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t = 8 s

Substituting into the equation,

v=0+(5)(8)=40 m/s

Learn more about accelerated motion:

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