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.
How much time passes before the car comes to rest?

Options

A. 5 sec
B. 4 sec
C. 8 sec
D. 6.5 sec

Answer

To solve the problem of how much time passes before the car comes to rest, we can follow these steps:

### Step 1: Understand the given data
- Mass of the car, \( m = 1000 \, \text{kg} \)
- Initial velocity of the car, \( v_0 = 25 \, \text{m/s} \)
- Rate of increase of braking force, \( \frac{dF}{dt} = 2000 \, \text{N/s} \)

### Step 2: Calculate the initial momentum
The initial momentum \( p_0 \) of the car can be calculated using the formula:
\[ p_0 = m \cdot v_0 = 1000 \, \text{kg} \cdot 25 \, \text{m/s} = 25000 \, \text{kg m/s} \]

### Step 3: Set up the impulse-momentum theorem
According to the impulse-momentum theorem:
\[ \text{Impulse} = \Delta p = p_f - p_0 \]
where \( p_f \) is the final momentum (which is 0 when the car comes to rest). Therefore:
\[ \Delta p = 0 - 25000 = -25000 \, \text{kg m/s} \]

### Step 4: Express the impulse in terms of force
The impulse can also be expressed as:
\[ \text{Impulse} = \int_0^{t} F(t) \, dt \]
Since the braking force \( F(t) \) increases linearly with time, we can express it as:
\[ F(t) = 2000 \, t \, \text{N} \]
Thus, the impulse becomes:
\[ \int_0^{t} F(t) \, dt = \int_0^{t} 2000 \, t \, dt = 2000 \int_0^{t} t \, dt = 2000 \left[ \frac{t^2}{2} \right]_0^{t} = 1000 t^2 \]

### Step 5: Set up the equation
Now we can set the impulse equal to the change in momentum:
\[ 1000 t^2 = 25000 \]

### Step 6: Solve for time \( t \)
Dividing both sides by 1000:
\[ t^2 = 25 \]
Taking the square root:
\[ t = 5 \, \text{s} \]

### Conclusion
The time that passes before the car comes to rest is \( t = 5 \) seconds.

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