Question

A car and a motor cycle start their motion from the same point with initial velocities \(23.7\text{ ms}^{-1}\) and zero respectively. If the accelerations of the motor cycle and car are \(3.8\text{ ms}^{-2}\) and zero respectively, then their relative velocity when the motor cycle crosses the car is:

• A. \(11.85\text{ ms}^{-1}\)
• B. \(23.7\text{ ms}^{-1}\)
• C. \(47.4\text{ ms}^{-2}\)
• D. \(47.4\text{ ms}^{-1}\)

Hint:

At the meeting point, first equate displacements to find time and then subtract velocities to obtain relative velocity.

Answer 1

The Correct Option is B

Concept: When two bodies move from the same point, they will meet again when their displacements become equal. Relative velocity is the difference between their velocities at that instant.

Step 1: Write displacement equations. For the car: \[ u_c=23.7\text{ ms}^{-1} \] \[ a_c=0 \] Hence \[ s_c=23.7t \] For the motorcycle: \[ u_m=0 \] \[ a_m=3.8\text{ ms}^{-2} \] Therefore \[ s_m=\frac12(3.8)t^2 \] \[ s_m=1.9t^2 \]

Step 2: Find the time when the motorcycle catches the car. At crossing, \[ s_c=s_m \] \[ 23.7t=1.9t^2 \] \[ t(1.9t-23.7)=0 \] Ignoring \(t=0\), \[ t=\frac{23.7}{1.9} \] \[ t\approx12.47\text{ s} \]

Step 3: Calculate motorcycle velocity. \[ v_m=u+at \] \[ v_m=0+3.8(12.47) \] \[ v_m\approx47.4\text{ ms}^{-1} \]

Step 4: Calculate relative velocity. Car velocity remains constant: \[ v_c=23.7\text{ ms}^{-1} \] Hence relative velocity \[ v_{rel}=v_m-v_c \] \[ v_{rel}=47.4-23.7 \] \[ v_{rel}=23.7\text{ ms}^{-1} \]

Step 5: Final answer. \[ \boxed{23.7\text{ ms}^{-1}} \]