The car's speed was zero at the beginning of the 12 seconds,
and 18 m/s at the end of it. Since the acceleration was 'uniform'
during that time, the car's average speed was (1/2)(0 + 18) = 9 m/s.
12 seconds at an average speed of 9 m/s ==> (12 x 9) = 108 meters .
==========================================
That's the way I like to brain it out. If you prefer to use the formula,
the first problem you run into is: You need to remember the formula !
The formula is D = 1/2 a T²
Distance = (1/2 acceleration) x (time in seconds)²
Acceleration = (change in speed) / (time for the change)
= (18 m/s) / (12 sec)
= 1.5 m/s² .
Distance = (1/2 x 1.5 m/s²) x (12 sec)²
= (0.75 m/s²) x (144 sec²) = 108 meters .
Answer:
so his far point according to this pair of glass is 200 cm
Explanation:
power of old pair of corrective glasses is given as
![P = -0.5 dioptre](https://tex.z-dn.net/?f=P%20%3D%20-0.5%20dioptre)
now we have
![f = \frac{1}{P}](https://tex.z-dn.net/?f=f%20%3D%20%5Cfrac%7B1%7D%7BP%7D)
![f = -2 m](https://tex.z-dn.net/?f=f%20%3D%20-2%20m)
![f = -200 cm](https://tex.z-dn.net/?f=f%20%3D%20-200%20cm)
now we know that for normal vision the maximum distance of vision is for infinite distance
so by lens formula we have
![\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bd_i%7D%20%2B%20%5Cfrac%7B1%7D%7Bd_o%7D%20%3D%20%5Cfrac%7B1%7D%7Bf%7D)
![\frac{1}{d_i} + 0 = \frac{1}{200}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bd_i%7D%20%2B%200%20%3D%20%5Cfrac%7B1%7D%7B200%7D)
![d_i = 200 cm](https://tex.z-dn.net/?f=d_i%20%3D%20200%20cm)
so his far point according to this pair of glass is 200 cm
Answer:
a) 0.462 m/s^2
b) 31.5 rad/s
c) 381 rad
d) 135m
Explanation:
the linear acceleration is given by:
![a=\alpha *r\\a=1.30rad/s^2*(35.5*10^{-2}m)\\a=0.462m/s^2](https://tex.z-dn.net/?f=a%3D%5Calpha%20%2Ar%5C%5Ca%3D1.30rad%2Fs%5E2%2A%2835.5%2A10%5E%7B-2%7Dm%29%5C%5Ca%3D0.462m%2Fs%5E2)
the angular speed is given by:
![\omega=\frac{v}{r}\\\\\omega=\frac{11.2m/s}{35.5*10^{-2}m}\\\\\omega=31.5rad/s](https://tex.z-dn.net/?f=%5Comega%3D%5Cfrac%7Bv%7D%7Br%7D%5C%5C%5C%5C%5Comega%3D%5Cfrac%7B11.2m%2Fs%7D%7B35.5%2A10%5E%7B-2%7Dm%7D%5C%5C%5C%5C%5Comega%3D31.5rad%2Fs)
to calculate how many radians have the wheel turned we need the apply the following formula:
![\theta=\frac{1}{2}\alpha*t^2\\\\t=\frac{\omega}{\alpha}\\\\t=\frac{31.5rad/s}{1.30rad/s^2}\\\\t=24.2s\\\\\theta=\frac{1}{2}*1.30rad/s^2*(24.2s)^2\\\\\theta=381rad](https://tex.z-dn.net/?f=%5Ctheta%3D%5Cfrac%7B1%7D%7B2%7D%5Calpha%2At%5E2%5C%5C%5C%5Ct%3D%5Cfrac%7B%5Comega%7D%7B%5Calpha%7D%5C%5C%5C%5Ct%3D%5Cfrac%7B31.5rad%2Fs%7D%7B1.30rad%2Fs%5E2%7D%5C%5C%5C%5Ct%3D24.2s%5C%5C%5C%5C%5Ctheta%3D%5Cfrac%7B1%7D%7B2%7D%2A1.30rad%2Fs%5E2%2A%2824.2s%29%5E2%5C%5C%5C%5C%5Ctheta%3D381rad)
the distance is given by:
![d=\theta*r](https://tex.z-dn.net/?f=d%3D%5Ctheta%2Ar)
![d=381rad*(35.5*10^{-2}m)\\d=135m](https://tex.z-dn.net/?f=d%3D381rad%2A%2835.5%2A10%5E%7B-2%7Dm%29%5C%5Cd%3D135m)
Answer:
Headlands and bays are created where there are bands of hard and soft rock which meet the coastline at right angles. Softer rock is eroded more quickly and erodes backwards to form bays (which may have beaches). The harder rocks are more resistant to erosion and jut out into the sea to form exposed headlands