To solve the problem, we need to first convert the initial velocity from km/h to m/s, as the other measurements (distance, mass) are in the metric system.
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Convert velocity: \[ \text{Initial velocity } (u) = 90 \text{ km/h} = \frac{90 \times 1000}{3600} \text{ m/s} = 25 \text{ m/s} \]
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Calculate the deceleration: We can use the kinematic equation: \[ v^2 = u^2 + 2as \] where
- \( v \) is the final velocity (0 m/s, since the car comes to rest),
- \( u \) is the initial velocity (25 m/s),
- \( a \) is the acceleration (deceleration in this case, so it will be negative),
- \( s \) is the distance (60 m).
Plugging in the values: \[ 0 = (25)^2 + 2a(60) \] Simplifying this gives: \[ 0 = 625 + 120a \] Solving for \( a \): \[ 120a = -625 \implies a = -\frac{625}{120} \approx -5.21 \text{ m/s}^2 \] Thus, the deceleration is approximately 5.21 m/s².
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Calculate the braking force: We can use Newton's second law: \[ F = ma \] where
- \( m \) is the mass of the car (900 kg),
- \( a \) is the deceleration calculated earlier.
Plugging in the values: \[ F = 900 \times (-5.21) \approx -4689 \text{ N} \] The negative sign indicates that the force is acting in the opposite direction of the motion. The braking force is approximately 4689 N.
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Calculate the time taken to bring the car to rest: We can use the kinematic equation: \[ v = u + at \] Solving for \( t \): \[ 0 = 25 + (-5.21)t \] This simplifies to: \[ 5.21t = 25 \implies t = \frac{25}{5.21} \approx 4.8 \text{ s} \] Thus, the time taken to bring the car to rest is approximately 4.8 seconds.
Summary:
- Deceleration: \( \approx 5.21 \text{ m/s}^2 \)
- Braking Force: \( \approx 4689 \text{ N} \)
- Time to bring to rest: \( \approx 4.8 \text{ s} \)