Question

A car is moving with an initial speed of 5 m/s. A constant braking force is applied and the car is brought to rest in a distance of 10 m. What is the average speed of the car during the deceleration process?

• A. \(\frac{20}{3}m\)
• B. 180 m
• C. 60 m
• D. 2.5 m
• E. 7 m

Answer 1

The Correct Option is D

The average speed of an object is defined as the total distance traveled divided by the total time taken. In this case, the car travels a distance of 10 m. We need to find the time it takes for the car to come to rest. 

We can use the following equations of motion: 

1. $v_f = v_i + at$ 2. $d = v_i t + \frac{1}{2}at^2$ 3. $v_f^2 = v_i^2 + 2ad$ 

Here, $v_i = 5$ m/s (initial speed), $v_f = 0$ m/s (final speed), $d = 10$ m (distance), and we need to find $t$ (time). 

We can use equation (3) to find the acceleration: $0^2 = 5^2 + 2a(10)$ $0 = 25 + 20a$ $20a = -25$ $a = -\frac{25}{20} = -\frac{5}{4}$ m/s$^2$ 

Now we can use equation (1) to find the time: $0 = 5 + (-\frac{5}{4})t$ $\frac{5}{4}t = 5$ $t = 5 \times \frac{4}{5} = 4$ s 

Now we can calculate the average speed: Average speed $= \frac{\text{total distance}}{\text{total time}} = \frac{10 \text{ m}}{4 \text{ s}} = 2.5$ m/s.

Final Answer: The final answer is $\boxed{B}$

Answer 2

Step 1: Recall the formula for average speed.

The average speed of an object is given by:

\[ \text{Average speed} = \frac{\text{Total distance traveled}}{\text{Total time taken}}. \]

We are given:

• Initial speed (\( u \)) = \( 5 \, \text{m/s} \),
• Final speed (\( v \)) = \( 0 \, \text{m/s} \) (since the car comes to rest),
• Distance traveled (\( s \)) = \( 10 \, \text{m} \).

 

We need to calculate the total time taken (\( t \)) to find the average speed.

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Step 2: Use the equations of motion to find the time taken.

From the equations of motion, we know:

\[ v^2 = u^2 + 2as, \]

where \( a \) is the acceleration (or deceleration in this case).

Substitute the known values (\( v = 0 \), \( u = 5 \, \text{m/s} \), \( s = 10 \, \text{m} \)):

\[ 0^2 = 5^2 + 2a(10). \]

Simplify:

\[ 0 = 25 + 20a. \]

Solve for \( a \):

\[ a = -\frac{25}{20} = -1.25 \, \text{m/s}^2. \]

The negative sign indicates deceleration. Now use another equation of motion to find the time taken:

\[ v = u + at. \]

Substitute \( v = 0 \), \( u = 5 \, \text{m/s} \), and \( a = -1.25 \, \text{m/s}^2 \):

\[ 0 = 5 + (-1.25)t. \]

Solve for \( t \):

\[ t = \frac{5}{1.25} = 4 \, \text{s}. \] ---

Step 3: Calculate the average speed.

Now substitute the total distance (\( s = 10 \, \text{m} \)) and total time (\( t = 4 \, \text{s} \)) into the formula for average speed:

\[ \text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{10}{4} = 2.5 \, \text{m/s}. \] ---

Final Answer: The average speed of the car during the deceleration process is \( \mathbf{2.5 \, \text{m/s}} \), which corresponds to option \( \mathbf{(B)} \).