Questions: A car going down a (20^circ) hill speeds up from (30 mathrm~m / mathrms) with an acceleration of (3 mathrm~m / mathrms^2) for 6 seconds. The car has a mass of 1000 kg , friction is negligible ((mathrmk=0)). Find how far the car went after those 6 seconds, in meters. [ cos (20^circ)=0.9397 sin (20^circ)=0.3420 ]

Solution Steps

We are given the following information:

• Initial velocity, \( v_0 = 30 \, \text{m/s} \)
• Acceleration, \( a = 3 \, \text{m/s}^2 \)
• Time, \( t = 6 \, \text{s} \)
• Mass of the car, \( m = 1000 \, \text{kg} \) (not needed for this calculation)
• Friction is negligible

To find the distance traveled by the car, we use the kinematic equation for uniformly accelerated motion:

\[ s = v_0 t + \frac{1}{2} a t^2 \]

where:

• \( s \) is the distance traveled,
• \( v_0 \) is the initial velocity,
• \( a \) is the acceleration,
• \( t \) is the time.

Substitute the given values into the kinematic equation:

\[ s = (30 \, \text{m/s}) \times (6 \, \text{s}) + \frac{1}{2} \times (3 \, \text{m/s}^2) \times (6 \, \text{s})^2 \]

Calculate each term:

1. \( v_0 t = 30 \times 6 = 180 \, \text{m} \)
2. \( \frac{1}{2} a t^2 = \frac{1}{2} \times 3 \times 36 = 54 \, \text{m} \)

Add these to find the total distance:

\[ s = 180 \, \text{m} + 54 \, \text{m} = 234 \, \text{m} \]

Final Answer