Question

A bus moving along a straight highway with speed of 72 km/h is brought to halt within 4s after applying the brakes. The distance travelled by the bus during this time (Assume the retardation is uniform) is _______m.

Answer 1

Correct Answer: 40

1. Convert Initial Velocity to m/s:
The initial velocity \( u = 72 \, \text{km/h} \) can be converted to m/s:
\[ u = 72 \times \frac{1000}{3600} = 20 \, \text{m/s}. \]

2. Use Equation of Motion to Find Retardation:
Using \( v = u + at \) with final velocity \( v = 0 \), time \( t = 4 \, \text{s} \):
\[ 0 = 20 + a \times 4. \] Solving for \( a \):
\[ a = -5 \, \text{m/s}^2. \] 

3. Calculate Distance Using \( v^2 - u^2 = 2as \):
Substitute values:
\[ 0^2 - 20^2 = 2 \times (-5) \times s. \] Simplifying:
\[ s = 40 \, \text{m}. \] 

Answer: 40 m

Answer 2

Step 1: Write the given data
Initial speed of the bus, \( u = 72 \, \text{km/h} \)
Time to come to rest, \( t = 4 \, \text{s} \)
Final speed, \( v = 0 \, \text{m/s} \)
We are asked to find the distance travelled \( s \) during this time, assuming uniform retardation.

Step 2: Convert the initial speed into m/s
\[ u = 72 \times \frac{1000}{3600} = 20 \, \text{m/s} \]

Step 3: Use the equation of motion
For uniformly accelerated (or retarded) motion:
\[ v = u + a t \] Substitute \( v = 0 \):
\[ 0 = 20 + a(4) \] \[ a = -5 \, \text{m/s}^2 \] The negative sign indicates retardation (deceleration).

Step 4: Find the distance travelled
Using the equation:
\[ s = ut + \frac{1}{2} a t^2 \] Substitute the known values:
\[ s = (20)(4) + \frac{1}{2}(-5)(4)^2 \] \[ s = 80 - \frac{1}{2}(5)(16) \] \[ s = 80 - 40 = 40 \, \text{m} \]

Step 5: Interpretation
The bus travels 40 meters before coming to rest. The uniform retardation ensures a straight-line decrease in speed from 20 m/s to 0 over 4 seconds.

Final Answer:

40