Solve it in English

Here's the solution to your problem: a) Sketch a speed-time graph to represent the motion of the car from A to B. A speed-time graph would show speed (in m s^-1) on the y-axis and time (in s) on the x-axis. The motion consists of four segments: • Segment 1 (Constant Speed): The car moves at 20 m s^-1 for 25 s. This is represented by a horizontal line from (0, 20) to (25, 20). • Segment 2 (Deceleration): The car decelerates from 20 m s^-1 to 8 m s^-1 at 0.4 m s^-2. The duration of this segment is calculated in part (b) as 30 s. This is represented by a straight line sloping downwards from (25, 20) to (25+30, 8) = (55, 8). • Segment 3 (Constant Speed): The car moves at 8 m s^-1 for 60 s. This is represented by a horizontal line from (55, 8) to (55+60, 8) = (115, 8). • Segment 4 (Acceleration): The car accelerates from 8 m s^-1 to 20 m s^-1. The duration of this segment is calculated in part (c) as 40 s. This is represented by a straight line sloping upwards from (115, 8) to (115+40, 20) = (155, 20). b) Find the time for which the car is decelerating. Step 1: Identify the initial speed (u), final speed (v), and acceleration (a) for the deceleration phase. The initial speed is u = 20 m s^-1. The final speed is v = 8 m s^-1. The deceleration is 0.4 m s^-2, so the acceleration is a = -0.4 m s^-2. Step 2: Use the equation of motion v = u + at to find the time t. 8 = 20 + (-0.4)t Step 3: Solve for t. 8 - 20 &= -0.4t \\ -12 &= -0.4t \\ t &= (-12)/(-0.4) \\ t &= 30 s The time for which the car is decelerating is 30 s. c) Given that the distance from A to B is 1960 m, find the time taken for the car to move from A to B. Step 1: Calculate the distance covered in the first three segments. • Segment 1 (Constant speed 20 m s^-1 for 25 s): d_1 = speed × time = 20 m s^-1 × 25 s = 500 m • Segment 2 (Deceleration from 20 m s^-1 to 8 m s^-1 over 30 s): d_2 = ((u+v)/(2))t = (20 m s^-1 + 8 m s^-12) × 30 s = ((28)/(2)) × 30 s = 14 m s^-1 × 30 s = 420 m • Segment 3 (Constant speed 8 m s^-1 for 60 s): d_3 = speed × time = 8 m s^-1 × 60 s = 480 m Step 2: Calculate the distance covered in the fourth segment (d_4). The total distance is 1960 m. d_4 = 1960 m - (d_1 + d_2 + d_3) d_4 = 1960 m - (500 m + 420 m + 480 m) d_4 = 1960 m - 1400 m = 560 m Step 3: Calculate the time taken for the fourth segment (t_4). In this segment, the car accelerates from u = 8 m s^-1 to v = 20 m s^-1 over a distance d_4 = 560 m. Using the equation d = ((u+v)/(2))t: 560 m = (8 m s^-1 + 20 m s^-12)t_4 560 m = ((28)/(2))t_4 560 m = 14 m s^-1 × t_4 t_4 = 560 m14 m s^-1 = 40 s Step 4: Calculate the total time taken. Total time T = t_1 + t_2 + t_3 + t_4 T = 25 s + 30 s + 60 s + 40 s T = 155 s The total time taken for the car to move from A to B is 155 s. What's next? Send 'em!