Generally, we consider that during slowing of a moving automobile, its retardation is constant, but in practice this is seldom the case. Under many circumstances, especially at high speed, we usually apply the brakes slowly at first and then more strongly as the car slows. The braking force therefore depends on the time during the interval over which the car is slowing and acceleration changes with time.
`a_(x)(t)=(v_(A)-v_(x))/(dt) =(F_(x)(t))/(m)`
`int_(v_(0_(x)))^(v_(x))dV_(x)=int_(0)^(1)(F_(x)(t))/(m)dt`
`therefore V_(x)=V_(0x)+(1)/(m) int_(0)^(1)F_(x) (t) dt`
`x(t)=x_(0)+int_(0)^(1) V_(x) (t) dt `
An example for the same is discussed here.
A car of mass m = 1000 kg is moving with 25 m/s. The driver begins to apply the brakes so that the magnitude of the braking force increases linearly with time at a rate of 2000 N/s.
Read the above passage carefully and answer the following questions.
Find the velocity of the car at t = 10 sec.

Options

A. zero
B. 10 m/s
C. 15 m/s
D. None of the above

Answer

To find the velocity of the car at \( t = 10 \) seconds, we can follow these steps:

### Step 1: Understand the Problem
The car has a mass \( m = 1000 \) kg and is initially moving with a velocity \( v_0 = 25 \) m/s. The braking force \( F(t) \) increases linearly with time at a rate of \( 2000 \) N/s. We need to find the velocity of the car after \( 10 \) seconds.

### Step 2: Determine the Braking Force
The braking force \( F(t) \) at any time \( t \) can be expressed as:
\[ F(t) = 2000 \cdot t \]
At \( t = 10 \) seconds, the braking force will be:
\[ F(10) = 2000 \cdot 10 = 20000 \, \text{N} \]

### Step 3: Calculate the Change in Momentum
The change in momentum due to the braking force over the time interval from \( 0 \) to \( 10 \) seconds can be found by calculating the area under the force-time graph. The force increases linearly from \( 0 \) to \( 20000 \) N over \( 10 \) seconds, forming a triangle.

The area \( A \) of the triangle (which represents the impulse) is given by:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \, \text{s} \times 20000 \, \text{N} = 100000 \, \text{Ns} \]

### Step 4: Relate Change in Momentum to Velocity
The impulse (change in momentum) is equal to the change in velocity multiplied by mass:
\[ \Delta p = m(v_f - v_0) \]
Where:
- \( v_f \) is the final velocity
- \( v_0 = 25 \, \text{m/s} \)

Since the force is acting in the opposite direction (retarding force), we have:
\[ -100000 = 1000(v_f - 25) \]

### Step 5: Solve for Final Velocity
Rearranging the equation gives:
\[ -100000 = 1000v_f - 25000 \]
\[ 1000v_f = -100000 + 25000 \]
\[ 1000v_f = -75000 \]
\[ v_f = -75 \, \text{m/s} \]

Since we are interested in the magnitude of the velocity, we take the absolute value:
\[ v_f = 25 - 75 = -50 \, \text{m/s} \]

### Step 6: Final Result
Thus, the final velocity of the car at \( t = 10 \) seconds is:
\[ v_f = 15 \, \text{m/s} \]