Question:

A car accelerates from rest at a constant rate of 2 m/s² for a time interval of 10 seconds. At the end of this time interval, the car comes to a stop and remains stationary for 5 seconds. Finally, the car accelerates in the opposite direction at a constant rate of -1 m/s² until it reaches a final velocity of -10 m/s.

a) Calculate the displacement of the car during the first 10 seconds. b) Calculate the displacement of the car during the entire motion.

Assume the positive direction is taken to be the initial direction of motion.

Answer:

a) To calculate the displacement during the first 10 seconds, we need to determine the distance covered during acceleration and deceleration separately.

During acceleration: Using the equation of motion: v = u + at

Initial velocity (u) = 0 m/s (since the car starts from rest) Acceleration (a) = 2 m/s² Time (t) = 10 s

Substituting these values into the equation, we get: v = (0 m/s) + (2 m/s²)(10 s) v = 20 m/s

The distance covered during acceleration can be calculated using the equation of motion: s = ut + 0.5at²

Substituting the known values, we obtain: s_acceleration = (0 m/s)(10 s) + 0.5(2 m/s²)(10 s)² s_acceleration = 0 m + 0.5(2 m/s²)(100 s²) s_acceleration = 0 m + 0.5(200 m) s_acceleration = 0 m + 100 m s_acceleration = 100 m

During deceleration: The initial velocity during deceleration is 20 m/s (which is the final velocity achieved during acceleration), and the acceleration is -1 m/s². To find the time it takes to reach the final velocity of -10 m/s, we can use the equation of motion: v = u + at

Final velocity (v) = -10 m/s Acceleration (a) = -1 m/s² Initial velocity (u) = 20 m/s

Rearranging the equation, we have: t = (v - u) / a t = (-10 m/s - 20 m/s) / (-1 m/s²) t = -30 m/s / -1 m/s² t = 30 s

Using the equation of motion: s = ut + 0.5at², we can calculate the distance during deceleration: s_deceleration = (20 m/s)(30 s) + 0.5(-1 m/s²)(30 s)² s_deceleration = 600 m - 0.5(900 m²/s²) s_deceleration = 600 m - 450 m²/s² s_deceleration = 600 m - 450 m s_deceleration = 150 m

Therefore, the total distance covered during the first 10 seconds is: s_total = s_acceleration + s_deceleration s_total = 100 m + 150 m s_total = 250 m

b) To calculate the displacement for the entire motion, we need to account for the direction of motion as well.

Since the car comes to a stop after the first 10 seconds, and then accelerates in the opposite direction until reaching a final velocity of -10 m/s, the displacement is the net distance covered in the opposite direction during deceleration.

Therefore, the displacement is given by: displacement = distance during deceleration = s_deceleration = 150 m

Hence, the displacement of the car during the entire motion is 150 m.