Answer:
(a) Distance traveled = 75.3846 m
(b) Velocity of car at that instant will be 14 m/sec
Explanation:
We have given acceleration of the car ![a=1.3m/sec^2](https://tex.z-dn.net/?f=a%3D1.3m%2Fsec%5E2)
Initial velocity of the cart u = 0 m/sec
(a) According to second equation of motion we know that ![s=ut+\frac{1}{2}at^2](https://tex.z-dn.net/?f=s%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2)
So distance traveled by car ![s_c=0\times t+\frac{1}{2}\times 1.3t^2=0.65t^2](https://tex.z-dn.net/?f=s_c%3D0%5Ctimes%20t%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%201.3t%5E2%3D0.65t%5E2)
As the truck is moving with constant speed
So distance traveled by truck ![s_t=ut=7t](https://tex.z-dn.net/?f=s_t%3Dut%3D7t)
As the truck overtakes the car
So ![s_c=s_t](https://tex.z-dn.net/?f=s_c%3Ds_t)
![0.65t^2=7t](https://tex.z-dn.net/?f=0.65t%5E2%3D7t)
![t=10.769sec](https://tex.z-dn.net/?f=t%3D10.769sec)
So distance traveled ![s_c=s_t=7\times 10.769=75.3846m](https://tex.z-dn.net/?f=s_c%3Ds_t%3D7%5Ctimes%2010.769%3D75.3846m)
(b) From second equation of motion we know that v = u+at
So v = 0+1.3×10.769 = 14 m /sec
The correct answer
to this question would be:
<span><span>
A. </span>
No part of your vehicle will extend out into
the traffic lane.</span>
This kind of maneuver only shows your skill
to handle the vehicle in tight spaces, ability to judge distance, and showing control
of the vehicle as you turn into a straight-in parking space.
<span> </span>
1,4,6
a bow is drawn back
a gun is loaded w/ a dart
a bungee cord is stretched
Answer:
no it will not glow hope help you stay happy