Question

1. (a) a tape attached to an accelerating trolley passes through a ticker timer that makes dots on it at a frequency of 50hz. the ticker makes 10 dots on a 10cm long tape such that; the distance a between the first two dots is 0.5cm and the distance b between the last two dots is 1.5cm.

i) determine the velocity of the trolley at:
i: distance a
(4 mks)
ii: distance b
(2 mks)
ii) determine the acceleration of the trolley

Explanation:

Step1: Calculate the time interval between two consecutive dots

The frequency \( f = 50Hz \). The time - period \( T=\frac{1}{f} \). So \( T=\frac{1}{50}=0.02s \)

Step2: Calculate the velocity at distance \( a \) (using the formula \( v=\frac{d}{t} \), where \( d \) is the distance between two dots and \( t \) is the time interval between two dots)

For distance \( a = 0.5cm=0.005m \), \( t = T=0.02s \). The velocity \( v_{a}=\frac{a}{T} \)
\( v_{a}=\frac{0.005}{0.02}=0.25m/s \)

Step3: Calculate the velocity at distance \( b \)

For distance \( b = 1.5cm = 0.015m \), \( t = T = 0.02s \). The velocity \( v_{b}=\frac{b}{T} \)
\( v_{b}=\frac{0.015}{0.02}=0.75m/s \)

Step4: Calculate the acceleration (using the formula \( a=\frac{v_{b}-v_{a}}{9T} \), because there are 9 intervals between the first and the last dot)

\( a=\frac{0.75 - 0.25}{9\times0.02}=\frac{0.5}{0.18}\approx2.78m/s^{2} \)

Answer:

I: The velocity at distance \( a \) is \( 0.25m/s \)
II: The velocity at distance \( b \) is \( 0.75m/s \)
ii) The acceleration of the trolley is approximately \( 2.78m/s^{2} \)