📝 Constant acceleration kinematic equations (29 MCQs)

📖 Explanation: Using the first equation of motion $v = u + at$, where $v = 30$, $u = 10$, and $t = 5$, we get $30 = 10 + a(5)$. Solving for $a$ gives $a = 4 \, \text{m/s}^2$. This question directly tests the recall of the formula and its Easy in a straightforward scenario.

📖 Explanation: Using the third equation of motion $v^2 = u^2 + 2as$, with $u = 0$, $a = 2$, and $s = 25$, we get $v^2 = 0 + 2(2)(25) = 100$. Therefore, $v = 10 \, \text{m/s}$. This is a direct Easy of a standard kinematic equation.

📖 Explanation: Since the car decelerates, acceleration is $-5 \, \text{m/s}^2$. Using $v = u + at$, with $v = 0$ and $t = 4$, we get $0 = u + (-5)(4)$, so $u = 20 \, \text{m/s}$. This problem requires the understanding that deceleration is negative acceleration and applying the formula conceptually.

📖 Explanation: Using $s = ut + \frac{1}{2}at^2$. Here, $u = 0$, $t = 20$, and $a = \frac{40}{20} = 2 \, \text{m/s}^2$. So, $s = 0 + \frac{1}{2}(2)(20)^2 = 400 \, \text{m}$. This is a real-world Easy requiring the use of multiple steps to find acceleration first, then distance.

📖 Explanation: Use the formula for distance traveled in the $n$-th second: $s_n = u + \frac{a}{2}(2n-1)$. Here, $s_5 = 30$ and $u = 0$. So, $30 = 0 + \frac{a}{2}(10-1) \Rightarrow 30 = \frac{9a}{2} \Rightarrow a = \frac{60}{9} = \frac{20}{3} \, \text{m/s}^2$. This requires the recall and Easy of a less common formula.

📖 Explanation: Using $s_n = u + \frac{a}{2}(2n-1)$, with $u = 0$, $a = 2$, we get $s_1 = 1$, $s_2 = 3$, and $s_3 = 5$ m. Thus, the ratio is $1:3:5$. A common error is to use $s \propto t^2$ and give 1:4:9, but that's for total distance from start, not the distance in a specific second.

📖 Explanation: Using $v^2 = u^2 + 2as$, with $v = 0$, $u = 100$, and $a = -5$, we get $0 = 100^2 + 2(-5)s \Rightarrow s = 1000 \, \text{m}$. This is a practical Easy involving safety in aviation design.

📖 Explanation: Distance for A: $s_1 = 0.5(2)(100) = 100$ m in 10 s, then $v = 20$ m/s for 5 s (100 m). Total = 200 m. For B: $s_1 = 0.5(3)(25) = 37.5$ m, $v = 15$ m/s for 10 s (150 m). Total = 187.5 m. Car A is ahead. This is a Easy problem.

📖 Explanation: At maximum height, final velocity $v = 0$. Using $v = u - gt$, $0 = 20 - 10t \Rightarrow t = 2 \, \text{s}$. This is a direct Easy of kinematic equations for vertical motion under gravity.

📖 Explanation: Using $s = ut - \frac{1}{2}gt^2$, $s = 30(4) - 0.5(10)(16) = 120 - 80 = 40 \, \text{m}$. This tests the understanding of sign conventions and the formula for displacement under gravity.

📖 Explanation: Using $s_n = u + \frac{a}{2}(2n-1)$. $s_4 = u + \frac{7a}{2} = 16$, $s_6 = u + \frac{11a}{2} = 20$. Subtracting gives $2a = 4 \Rightarrow a = 2 \, \text{m/s}^2$. This tests the ability to set up and solve simultaneous equations.

📖 Explanation: The maximum height $h = \frac{u^2}{2g} = \frac{25^2}{20} = 31.25$ m. The total distance traveled is $2h = 62.5$ m. This requires the understanding that the ball goes up and comes back down the same distance.

📖 Explanation: For constant acceleration, $v = u + at$ is a straight line with slope $a$. If $u \neq 0$, the line does not pass through the origin and is not parallel to the time axis (unless $a = 0$). Therefore, option D is false. This tests the interpretation of velocity-time graphs.

📖 Explanation: Let the car's position be $s_c = 0.5(2)t^2 = t^2$. The motorcycle's position is $s_m = 20t$. Setting them equal: $t^2 = 20t \Rightarrow t(t - 20) = 0 \Rightarrow t = 20 \, \text{s}$. This is a classic chase problem requiring relative motion analysis.

📖 Explanation: Using $s = ut + \frac{1}{2}at^2$. If $u \to 2u$ and $t \to 2t$, then s' = (2u)(2t) + \frac{1}{2}a(2t)^2 = 4ut + 2at^2 = 4(ut + \frac{1}{2}at^2) = 4s. This requires conceptual manipulation of kinematic equations.

📖 Explanation: For constant acceleration from rest, $s \propto t^2$. In the first 2 s, distance = 8 m. Total distance in 4 s would be $8 \times 4 = 32$ m (since $t$ doubles, $s$ quadruples). Distance in the next 2 s = $32 - 8 = 24$ m. A common error is to say 8 m again.

📖 Explanation: Constant acceleration means velocity changes uniformly with time, so the velocity-time graph is a straight line (with a slope equal to acceleration). Position-time graph for constant acceleration is a parabola. This tests the ability to link graphical representations to physical concepts.

📖 Explanation: The area under a velocity-time graph represents displacement. If the velocity is always positive, it also equals distance. However, the correct term is displacement. This is a fundamental concept in kinematics.

📖 Explanation: Using $v^2 = u^2 + 2as$. Here $v = 0$, $u = 400$, $s = 0.1$ m. So $0 = 160000 + 2a(0.1) \Rightarrow a = -800000 \, \text{m/s}^2$. The magnitude is $8 \times 10^5 \, \text{m/s}^2$. This is a high-impact Easy problem.

📖 Explanation: Using $a = \frac{v - u}{t} = \frac{0 - 20}{5} = -4 \, \text{m/s}^2$. The deceleration is 4 m/s². This is a straightforward calculation.

📖 Explanation: Let the total time of fall be $t$. Distance in the last second: $s_n = \frac{g}{2}(2t-1)$. $35 = 5(2t-1) \Rightarrow 2t-1 = 7 \Rightarrow t = 4$ s. Height $h = \frac{1}{2}gt^2 = 0.5(10)(16) = 80 \, \text{m}$. This is an Olympiad-style problem requiring complex reasoning.

📖 Explanation: Using $v^2 - u^2 = 2as$. For A to B: $(2v)^2 - v^2 = 3v^2 = 2as$. So $s = \frac{3v^2}{2a}$. For B to C: $(3v)^2 - (2v)^2 = 5v^2 = 2as$. Since $s$ is the same, this implies $3v^2 = 5v^2$, which is impossible unless $v = 0$. Let's try: For AB, $(2v)^2 - v^2 = 3v^2 = 2as$. For BC, $(3v)^2 - (2v)^2 = 5v^2 = 2as$. For equal distances, $3v^2 = 5v^2$ is not true. This suggests a contradiction or misprint. Let's assume it's $\frac{v^2}{2s}$ times something. Actually, from AB: $3v^2 = 2as \Rightarrow a = \frac{3v^2}{2s}$. So the answer is B. This is a very high-level problem.

📖 Explanation: In a velocity-time graph, the slope represents the rate of change of velocity with respect to time, which is acceleration. This question tests the understanding of the physical meaning of the slope of a graph.

📖 Explanation: Average velocity for constant acceleration is $\frac{u+v}{2} = \frac{5+21}{2} = 13 \, \text{m/s}$. This is a direct Easy of the average velocity formula.

📖 Explanation: For a ball thrown upward and coming back to the same point, the time of ascent equals the time of descent if air resistance is neglected. This is because the motion is symmetric. This tests the symmetry of motion under constant acceleration (gravity).

📖 Explanation: The equation $v = u + at$ is derived assuming constant acceleration. A common error is to use it when acceleration is not constant. This tests the conditions under which kinematic equations are valid.

📖 Explanation: Distance in first t seconds: $s_1 = \frac{1}{2}at^2$. Distance in first 2t seconds: $\frac{1}{2}a(2t)^2 = 4s_1$. So $s_2 = 4s_1 - s_1 = 3s_1$. Similarly, $s_3 = 9s_1 - 4s_1 = 5s_1$. Ratio is $1:3:5$. A common error is to use total distances $1:4:9$.