A car initially at rest moves with a constant acceleration of 2m.s⁻² east. Calculate the Magnitude of the velocity after 10 s and Distance covered during the first 10 s.

Question

Asked on June 23, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-06-23T07:24:23.649Z

💍🫀Thakgiey, let's knock this out. QUESTION 4 A car initially at rest moves with a constant acceleration of 2 m/s^2 east. 4.1 Magnitude of the velocity after 10 s Step 1: Identify the given values. Initial velocity u = 0 m/s (at rest). Acceleration a = 2 m/s^2. Time t = 10 s. Step 2: Use the kinematic equation v = u + at to find the final velocity v. v = 0 m/s + (2 m/s^2)(10 s) v = 20 m/s The magnitude of the velocity after 10 s is 20 m/s. 4.2 Distance covered during the first 10 s Step 1: Identify the given values. Initial velocity u = 0 m/s. Acceleration a = 2 m/s^2. Time t = 10 s. Step 2: Use the kinematic equation s = ut + (1)/(2)at^2 to find the distance s. s = (0 m/s)(10 s) + (1)/(2)(2 m/s^2)(10 s)^2 s = 0 + (1)/(2)(2)(100) s = 100 m The distance covered during the first 10 s is 100 m. --- QUESTION 6 6.1 Define the term acceleration. Acceleration is the rate of change of velocity of an object over time. It is a vector quantity, meaning it has both magnitude and direction. 6.2 Convert 77 km.h⁻¹ to m.s⁻¹. Step 1: Convert kilometers to meters and hours to seconds. 1 km = 1000 m 1 h = 3600 s Step 2: Apply the conversion factors. 77 km/h = 77 × 1000 m3600 s 77 km/h = 77 × (5)/(18) m/s 77 km/h ≈ 21.39 m/s The velocity of car A is 21.39 m/s. 6.3 Determine the velocity of CAR B while chasing CAR A. Step 1: Identify the given values for car B. Initial velocity u = 0 m/s (from rest). Acceleration a = 1.5 m/s^2. Time t = 6.5 s. Step 2: Use the kinematic equation v = u + at to find the final velocity v. v = 0 m/s + (1.5 m/s^2)(6.5 s) v = 9.75 m/s The velocity of car B after 6.5 s is 9.75 m/s. 6.4 How far has CAR B driven after 6,5 seconds? Step 1: Identify the given values for car B. Initial velocity u = 0 m/s. Acceleration a = 1.5 m/s^2. Time t = 6.5 s. Step 2: Use the kinematic equation s = ut + (1)/(2)at^2 to find the distance s. s = (0 m/s)(6.5 s) + (1)/(2)(1.5 m/s^2)(6.5 s)^2 s = 0 + (1)/(2)(1.5)(42.25) s = 0.75 × 42.25 s = 31.6875 m Car B has driven 31.69 m (rounded to two decimal places). 6.5 How far did CAR A travel in that 6,5 seconds? Step 1: Identify the given values for car A. Constant velocity v_A = 77 km/h = 21.39 m/s (from 6.2). Time t = 6.5 s. Step 2: Use the formula for distance with constant velocity s = v × t. s_A = (21.39 m/s)(6.5 s) s_A = 139.035 m Car A traveled 139.04 m (rounded to two decimal places). 6.6 Will CAR B be able to catch up with CAR A? Explain your answer. Step 1: Compare the distances covered by both cars after 6.5 s. Car A's distance s_A = 139.04 m. Car B's distance s_B = 31.69 m. At 6.5 s, Car B is behind Car A (s_B < s_A). Step 2: Compare their velocities after 6.5 s. Car A's velocity v_A = 21.39 m/s (constant). Car B's velocity v_B = 9.75 m/s (after 6.5 s, it continues at this constant velocity). Since Car B's constant velocity (9.75 m/s) is less than Car A's constant velocity (21.39 m/s), Car B will not be able to catch up with Car A. Explanation: ❌ No, Car B will not be able to catch up with Car A. After 6.5 s, Car B is still behind Car A (31.69 m vs 139.04 m). Furthermore, Car B's velocity (9.75 m/s) is less than Car A's velocity (21.39 m/s), and Car B continues at this constant, slower velocity. Therefore, the distance between them will continue to increase. 6.7 What acceleration does CAR A experience to come to a standstill after 2,5 s? Step 1: Identify the given values for car A during braking. Final velocity v = 0 m/s (comes to a standstill). Distance s = 25 m. Time t = 2.5 s. Step 2: Use the kinematic equations to find acceleration a. We have s = ut + (1)/(2)at^2 and v = u + at. From v = u + at, we can express u as u = v - at = 0 - a(2.5) = -2.5a. Substitute this into the distance equation: s = (-2.5a)t + (1)/(2)at^2 25 m = (-2.5a)(2.5 s) + (1)/(2)a(2.5 s)^2 25 = -6.25a + (1)/(2)a(6.25) 25 = -6.25a + 3.125a 25 = -3.125a a = (25)/(-3.125) a = -8 m/s^2 The acceleration car A experiences is -8 m/s^2. The negative sign indicates deceleration. Drop the next question.

Accepted Answer · 2026-06-23T07:24:23.649Z

💍🫀Thakgiey, let's knock this out. QUESTION 4 A car initially at rest moves with a constant acceleration of 2 m/s^2 east. 4.1 Magnitude of the velocity after 10 s Step 1: Identify the given values. Initial velocity u = 0 m/s (at rest). Acceleration a = 2 m/s^2. Time t = 10 s. Step 2: Use the kinematic equation v = u + at to find the final velocity v. v = 0 m/s + (2 m/s^2)(10 s) v = 20 m/s The magnitude of the velocity after 10 s is 20 m/s. 4.2 Distance covered during the first 10 s Step 1: Identify the given values. Initial velocity u = 0 m/s. Acceleration a = 2 m/s^2. Time t = 10 s. Step 2: Use the kinematic equation s = ut + (1)/(2)at^2 to find the distance s. s = (0 m/s)(10 s) + (1)/(2)(2 m/s^2)(10 s)^2 s = 0 + (1)/(2)(2)(100) s = 100 m The distance covered during the first 10 s is 100 m. --- QUESTION 6 6.1 Define the term acceleration. Acceleration is the rate of change of velocity of an object over time. It is a vector quantity, meaning it has both magnitude and direction. 6.2 Convert 77 km.h⁻¹ to m.s⁻¹. Step 1: Convert kilometers to meters and hours to seconds. 1 km = 1000 m 1 h = 3600 s Step 2: Apply the conversion factors. 77 km/h = 77 × 1000 m3600 s 77 km/h = 77 × (5)/(18) m/s 77 km/h ≈ 21.39 m/s The velocity of car A is 21.39 m/s. 6.3 Determine the velocity of CAR B while chasing CAR A. Step 1: Identify the given values for car B. Initial velocity u = 0 m/s (from rest). Acceleration a = 1.5 m/s^2. Time t = 6.5 s. Step 2: Use the kinematic equation v = u + at to find the final velocity v. v = 0 m/s + (1.5 m/s^2)(6.5 s) v = 9.75 m/s The velocity of car B after 6.5 s is 9.75 m/s. 6.4 How far has CAR B driven after 6,5 seconds? Step 1: Identify the given values for car B. Initial velocity u = 0 m/s. Acceleration a = 1.5 m/s^2. Time t = 6.5 s. Step 2: Use the kinematic equation s = ut + (1)/(2)at^2 to find the distance s. s = (0 m/s)(6.5 s) + (1)/(2)(1.5 m/s^2)(6.5 s)^2 s = 0 + (1)/(2)(1.5)(42.25) s = 0.75 × 42.25 s = 31.6875 m Car B has driven 31.69 m (rounded to two decimal places). 6.5 How far did CAR A travel in that 6,5 seconds? Step 1: Identify the given values for car A. Constant velocity v_A = 77 km/h = 21.39 m/s (from 6.2). Time t = 6.5 s. Step 2: Use the formula for distance with constant velocity s = v × t. s_A = (21.39 m/s)(6.5 s) s_A = 139.035 m Car A traveled 139.04 m (rounded to two decimal places). 6.6 Will CAR B be able to catch up with CAR A? Explain your answer. Step 1: Compare the distances covered by both cars after 6.5 s. Car A's distance s_A = 139.04 m. Car B's distance s_B = 31.69 m. At 6.5 s, Car B is behind Car A (s_B < s_A). Step 2: Compare their velocities after 6.5 s. Car A's velocity v_A = 21.39 m/s (constant). Car B's velocity v_B = 9.75 m/s (after 6.5 s, it continues at this constant velocity). Since Car B's constant velocity (9.75 m/s) is less than Car A's constant velocity (21.39 m/s), Car B will not be able to catch up with Car A. Explanation: ❌ No, Car B will not be able to catch up with Car A. After 6.5 s, Car B is still behind Car A (31.69 m vs 139.04 m). Furthermore, Car B's velocity (9.75 m/s) is less than Car A's velocity (21.39 m/s), and Car B continues at this constant, slower velocity. Therefore, the distance between them will continue to increase. 6.7 What acceleration does CAR A experience to come to a standstill after 2,5 s? Step 1: Identify the given values for car A during braking. Final velocity v = 0 m/s (comes to a standstill). Distance s = 25 m. Time t = 2.5 s. Step 2: Use the kinematic equations to find acceleration a. We have s = ut + (1)/(2)at^2 and v = u + at. From v = u + at, we can express u as u = v - at = 0 - a(2.5) = -2.5a. Substitute this into the distance equation: s = (-2.5a)t + (1)/(2)at^2 25 m = (-2.5a)(2.5 s) + (1)/(2)a(2.5 s)^2 25 = -6.25a + (1)/(2)a(6.25) 25 = -6.25a + 3.125a 25 = -3.125a a = (25)/(-3.125) a = -8 m/s^2 The acceleration car A experiences is -8 m/s^2. The negative sign indicates deceleration. Drop the next question.

A car initially at rest moves with a constant acceleration of 2m.s⁻² east. Calculate the Magnitude of the velocity after 10 s and Distance covered during the first 10 s.

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A car initially at rest moves with a constant acceleration of 2m.s⁻² east. Calculate the Magnitude of the velocity after 10 s and Distance covered during the first 10 s.

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