A car starts from rest and travels for 7.0 s with a uniform acceleration of +3.0 m/s2. The driver then applies the brakes, causing a uniform acceleration of −2.5 m/s2. If the brakes are applied for 2.0 s, determine each of the following.
How far has the car gone?
To determine the distance the car has gone, we need to consider two parts of its motion: the period of acceleration and the period of deceleration.
During the period of acceleration:
Initial velocity (u) = 0 m/s (starting from rest)
Acceleration (a) = +3.0 m/s²
Time (t) = 7.0 s
Using the equation of motion:
d = ut + (1/2)at²,
where d is the distance covered by the car.
Plugging in the values:
d = 0 * 7.0 + (1/2) * 3.0 * (7.0)²
d = 0 + (1/2) * 3.0 * 49.0
d = 0 + 1.5 * 49.0
d = 73.5 m
During the period of deceleration:
The car is now moving with an initial velocity (u) after accelerating for 7.0 seconds, and it decelerates uniformly with an acceleration (a) of -2.5 m/s² for a time (t) of 2.0 s.
Using the same equation of motion:
d = ut + (1/2)at²
Plugging in the values:
d = (3.0 m/s * 7.0 s) + (1/2) * (-2.5 m/s²) * (2.0 s)²
d = 21.0 m - 1/2 * 2.5 m/s² * 4.0 s²
d = 21.0 m - 5.0 m/s² * 4.0 s²
d = 21.0 m - 20.0 m
d = 1.0 m
Therefore, the total distance covered by the car is the sum of the distances during acceleration and deceleration:
Total distance = 73.5 m + 1.0 m
Total distance = 74.5 m
Hence, the car has traveled a distance of 74.5 meters.