A car starts from rest and moves with uniform acceleration for time t then it moves with uniform speed of 60 km/h for time 3t and comes to rest. Find the average speed of car in its total journey.

Answer

To solve the problem step by step, we will break down the journey of the car into three parts: acceleration, constant speed, and deceleration.

### Step 1: Define the parameters
- The car starts from rest (initial velocity \( u = 0 \)).
- It accelerates for time \( t \) and then moves at a constant speed of \( 60 \, \text{km/h} \) for \( 3t \).
- Finally, it comes to rest.

### Step 2: Calculate the distance during acceleration (d1)
Using the formula for distance under uniform acceleration:
\[ d_1 = ut + \frac{1}{2} a t^2 \]
Since the car starts from rest, \( u = 0 \):
\[ d_1 = \frac{1}{2} a t^2 \]

### Step 3: Find the final velocity after acceleration
The final velocity \( v \) after accelerating for time \( t \) can be found using:
\[ v = u + at \]
Again, since \( u = 0 \):
\[ v = at \]

### Step 4: Relate the final velocity to the constant speed
After accelerating for time \( t \), the car moves at a constant speed of \( 60 \, \text{km/h} \). Therefore:
\[ at = 60 \, \text{km/h} \]
To convert \( 60 \, \text{km/h} \) to \( \text{m/s} \):
\[ 60 \, \text{km/h} = \frac{60 \times 1000}{3600} = 16.67 \, \text{m/s} \]
Thus, we have:
\[ at = 16.67 \]

### Step 5: Calculate the distance during constant speed (d2)
The distance \( d_2 \) covered during the constant speed phase is:
\[ d_2 = v \times t = 60 \, \text{km/h} \times 3t \]
Converting \( 60 \, \text{km/h} \) to \( \text{m/s} \):
\[ d_2 = 16.67 \times 3t = 50.01t \]

### Step 6: Calculate the total distance
The total distance \( D \) covered is:
\[ D = d_1 + d_2 \]
Substituting \( d_1 \) and \( d_2 \):
\[ D = \frac{1}{2} a t^2 + 50.01t \]

### Step 7: Calculate the total time
The total time \( T \) taken is:
\[ T = t + 3t = 4t \]

### Step 8: Calculate the average speed
The average speed \( V_{avg} \) is given by:
\[ V_{avg} = \frac{D}{T} \]
Substituting \( D \) and \( T \):
\[ V_{avg} = \frac{\frac{1}{2} a t^2 + 50.01t}{4t} \]

### Step 9: Substitute \( a \)
From \( at = 16.67 \):
\[ a = \frac{16.67}{t} \]
Substituting \( a \) into the average speed formula:
\[ V_{avg} = \frac{\frac{1}{2} \left(\frac{16.67}{t}\right) t^2 + 50.01t}{4t} \]
\[ = \frac{8.335t + 50.01t}{4t} \]
\[ = \frac{58.345t}{4t} = 14.58625 \, \text{m/s} \]

### Step 10: Convert average speed to km/h
To convert \( V_{avg} \) back to km/h:
\[ V_{avg} = 14.58625 \times \frac{3600}{1000} = 52.5 \, \text{km/h} \]

### Final Answer
The average speed of the car in its total journey is \( 52.5 \, \text{km/h} \).

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