Given:
Initial speed of the motorcycle (u) = 35 m/s
Final speed of the motorcycle (v) = 0 m/s (Complete Stop)
Maximum deceleration of the motorcycle (a) = -1.2 m/s²
Required Equation:
![\boxed{\bf{ v = u + at}}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%5Cbf%7B%20v%20%3D%20u%20%2B%20at%7D%7D)
Answer:
By substituting values in the equation, we get:
![\rm \longrightarrow 0 = 35 + ( - 1.2)t \\ \\ \rm \longrightarrow 0 = 35 - 1.2t \\ \\ \rm \longrightarrow 35 - 1.2t = 0 \\ \\ \rm \longrightarrow 35- 35 - 1.2t = 0 - 35 \\ \\ \rm \longrightarrow - 1.2t = - 35 \\ \\ \rm \longrightarrow \dfrac{ - 1.2t}{ - 1.2} = \dfrac{ - 35}{ - 1.2} \\ \\ \rm \longrightarrow t = 29.167 \: s](https://tex.z-dn.net/?f=%20%5Crm%20%5Clongrightarrow%200%20%3D%2035%20%2B%20%28%20-%201.2%29t%20%5C%5C%20%20%5C%5C%20%20%5Crm%20%5Clongrightarrow%20%200%20%3D%2035%20-%201.2t%20%5C%5C%20%20%5C%5C%20%20%5Crm%20%20%5Clongrightarrow%2035%20-%201.2t%20%3D%200%20%5C%5C%20%20%5C%5C%20%20%5Crm%20%20%5Clongrightarrow%2035-%2035%20-%201.2t%20%3D%200%20-%2035%20%5C%5C%20%20%5C%5C%20%20%5Crm%20%20%5Clongrightarrow%20%20-%201.2t%20%3D%20%20-%2035%20%5C%5C%20%20%5C%5C%20%20%5Crm%20%5Clongrightarrow%20%20%5Cdfrac%7B%20-%201.2t%7D%7B%20-%201.2%7D%20%20%3D%20%20%5Cdfrac%7B%20-%2035%7D%7B%20-%201.2%7D%20%20%5C%5C%20%20%5C%5C%20%20%5Crm%20%5Clongrightarrow%20%20t%20%3D%2029.167%20%5C%3A%20s)
Time taken by motorcycle to come to a complete stop (t) = 29.167 s
The net force on the acorn is less than the force of gravity.
<h3><u>Answer;</u></h3>
A. 4
<h3><u>Explanation;</u></h3>
- <em><u>The period of a wave or periodic time is the time taken for a complete oscillation to occur. </u></em>For example its is the time taken between two successive crests or troughs.
- <em><u>The beats or oscillation that occur in one second represents the frequency. Frequency is the number of complete oscillations or beats in one second in a wave.</u></em>
- Frequency, measured in Hertz is given by the reciprocal of the periodic time.
- Thus; <u><em>Frequency or beats per second = 1/(1/4) = 4</em></u>
- <u><em>Hence , 4 beats per second</em></u>
Answer:
Explanation:
Given
Initial velocity of ball ![u=10\ m/s](https://tex.z-dn.net/?f=u%3D10%5C%20m%2Fs)
height of window ![h=20\ m](https://tex.z-dn.net/?f=h%3D20%5C%20m)
Using Equation of motion
![y=ut+\frac{1}{2}at^2](https://tex.z-dn.net/?f=y%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2)
where u=initial velocity
t=time
a=acceleration
As ball is already is at a height of 20 m so
![Y=ut+\frac{1}{2}at^2+20](https://tex.z-dn.net/?f=Y%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2%2B20)
![Y=10\times t+0.5\times (-9.8)t^2+20](https://tex.z-dn.net/?f=Y%3D10%5Ctimes%20t%2B0.5%5Ctimes%20%28-9.8%29t%5E2%2B20)
![Y=-4.9t^2+10t+20](https://tex.z-dn.net/?f=Y%3D-4.9t%5E2%2B10t%2B20)
(b)highest point is obtained at v=0
![v^2-u^2=2as](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as)
where
v=final velocity
u=initial velocity
a=acceleration
s=displacement
![(0)-10^2=2\times (-9.8)\times s](https://tex.z-dn.net/?f=%280%29-10%5E2%3D2%5Ctimes%20%28-9.8%29%5Ctimes%20s)
![s=\frac{100}{19.6}](https://tex.z-dn.net/?f=s%3D%5Cfrac%7B100%7D%7B19.6%7D)
![s=5.102\ m](https://tex.z-dn.net/?f=s%3D5.102%5C%20m)
Highest Point will be ![s+20=25.102\ m](https://tex.z-dn.net/?f=s%2B20%3D25.102%5C%20m)
(c)Time taken when the ball hit the ground i.e. at Y=0
![-4.9t^2+10t+20=0](https://tex.z-dn.net/?f=-4.9t%5E2%2B10t%2B20%3D0)
![t=3.28\ s](https://tex.z-dn.net/?f=t%3D3.28%5C%20s)
impact velocity ![v=\sqrt{2\times 9.8\times 25.102}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B2%5Ctimes%209.8%5Ctimes%2025.102%7D)