Top 50 Most Repeated MOTION IN A STRAIGHT LINE PYQs | JEE MAINS
A curated collection of the most important questions from MOTION IN A STRAIGHT LINE, fully solved with step-by-step concepts to prepare for JEE MAINS.
A curated collection of the most important questions from MOTION IN A STRAIGHT LINE, fully solved with step-by-step concepts to prepare for JEE MAINS.
The deceleration a = Δv/Δt = (0 − 20)/4 = −5 m/s². Using s = ut + ½ a t² gives s = 20×4 + ½(−5)×4² = 80 − 40 = 40 m. The other options arise from sign...
Read Full Step-by-Step Solution →Terminal velocity occurs when drag force equals weight, leading to $mg = kv_t$....
Read Full Step-by-Step Solution →Integrate $v$ wrt $t$: $\int_0^2 3t^2 dt = [t^3]_0^2 = 8$....
Read Full Step-by-Step Solution →Using $v^2 = u^2 + 2gh$, $v = \sqrt{2 \times 10 \times 80} = 40\,m/s$...
Read Full Step-by-Step Solution →Area under v-t graph gives displacement in the time interval....
Read Full Step-by-Step Solution →Using $v = u + at$, final velocity is $0 + 5 \times 10 = 50\,\text{m/s}$....
Read Full Step-by-Step Solution →Terminal velocity $v_t \propto r^2$ by Stokes' law....
Read Full Step-by-Step Solution →We can use the equation v = u + αt, where v is the final velocity, u is the initial velocity (0 m/s), α is the acceleration, and t is the time. Pluggi...
Read Full Step-by-Step Solution →The deceleration a = Δv/Δt = (0 − 20)/4 = −5 m/s². Using s = ut + ½ a t² gives s = 20×4 + ½(−5)×4² = 80 − 40 = 40 m. The other options arise from sign...
Read Full Step-by-Step Solution →Terminal velocity occurs when drag force equals weight, leading to $mg = kv_t$....
Read Full Step-by-Step Solution →Integrate $v$ wrt $t$: $\int_0^2 3t^2 dt = [t^3]_0^2 = 8$....
Read Full Step-by-Step Solution →Using $v^2 = u^2 + 2gh$, $v = \sqrt{2 \times 10 \times 80} = 40\,m/s$...
Read Full Step-by-Step Solution →Area under v-t graph gives displacement in the time interval....
Read Full Step-by-Step Solution →Using $v = u + at$, final velocity is $0 + 5 \times 10 = 50\,\text{m/s}$....
Read Full Step-by-Step Solution →Terminal velocity $v_t \propto r^2$ by Stokes' law....
Read Full Step-by-Step Solution →We can use the equation v = u + αt, where v is the final velocity, u is the initial velocity (0 m/s), α is the acceleration, and t is the time. Pluggi...
Read Full Step-by-Step Solution →The deceleration a = Δv/Δt = (0 − 20)/4 = −5 m/s². Using s = ut + ½ a t² gives s = 20×4 + ½(−5)×4² = 80 − 40 = 40 m. The other options arise from sign...
Read Full Step-by-Step Solution →Terminal velocity occurs when drag force equals weight, leading to $mg = kv_t$....
Read Full Step-by-Step Solution →Integrate $v$ wrt $t$: $\int_0^2 3t^2 dt = [t^3]_0^2 = 8$....
Read Full Step-by-Step Solution →Using $v^2 = u^2 + 2gh$, $v = \sqrt{2 \times 10 \times 80} = 40\,m/s$...
Read Full Step-by-Step Solution →Area under v-t graph gives displacement in the time interval....
Read Full Step-by-Step Solution →Using $v = u + at$, final velocity is $0 + 5 \times 10 = 50\,\text{m/s}$....
Read Full Step-by-Step Solution →Terminal velocity $v_t \propto r^2$ by Stokes' law....
Read Full Step-by-Step Solution →We can use the equation v = u + αt, where v is the final velocity, u is the initial velocity (0 m/s), α is the acceleration, and t is the time. Pluggi...
Read Full Step-by-Step Solution →The deceleration a = Δv/Δt = (0 − 20)/4 = −5 m/s². Using s = ut + ½ a t² gives s = 20×4 + ½(−5)×4² = 80 − 40 = 40 m. The other options arise from sign...
Read Full Step-by-Step Solution →Terminal velocity occurs when drag force equals weight, leading to $mg = kv_t$....
Read Full Step-by-Step Solution →Integrate $v$ wrt $t$: $\int_0^2 3t^2 dt = [t^3]_0^2 = 8$....
Read Full Step-by-Step Solution →Using $v^2 = u^2 + 2gh$, $v = \sqrt{2 \times 10 \times 80} = 40\,m/s$...
Read Full Step-by-Step Solution →Area under v-t graph gives displacement in the time interval....
Read Full Step-by-Step Solution →Using $v = u + at$, final velocity is $0 + 5 \times 10 = 50\,\text{m/s}$....
Read Full Step-by-Step Solution →Terminal velocity $v_t \propto r^2$ by Stokes' law....
Read Full Step-by-Step Solution →We can use the equation v = u + αt, where v is the final velocity, u is the initial velocity (0 m/s), α is the acceleration, and t is the time. Pluggi...
Read Full Step-by-Step Solution →The deceleration a = Δv/Δt = (0 − 20)/4 = −5 m/s². Using s = ut + ½ a t² gives s = 20×4 + ½(−5)×4² = 80 − 40 = 40 m. The other options arise from sign...
Read Full Step-by-Step Solution →Terminal velocity occurs when drag force equals weight, leading to $mg = kv_t$....
Read Full Step-by-Step Solution →Integrate $v$ wrt $t$: $\int_0^2 3t^2 dt = [t^3]_0^2 = 8$....
Read Full Step-by-Step Solution →Using $v^2 = u^2 + 2gh$, $v = \sqrt{2 \times 10 \times 80} = 40\,m/s$...
Read Full Step-by-Step Solution →Area under v-t graph gives displacement in the time interval....
Read Full Step-by-Step Solution →Using $v = u + at$, final velocity is $0 + 5 \times 10 = 50\,\text{m/s}$....
Read Full Step-by-Step Solution →Terminal velocity $v_t \propto r^2$ by Stokes' law....
Read Full Step-by-Step Solution →We can use the equation v = u + αt, where v is the final velocity, u is the initial velocity (0 m/s), α is the acceleration, and t is the time. Pluggi...
Read Full Step-by-Step Solution →The deceleration a = Δv/Δt = (0 − 20)/4 = −5 m/s². Using s = ut + ½ a t² gives s = 20×4 + ½(−5)×4² = 80 − 40 = 40 m. The other options arise from sign...
Read Full Step-by-Step Solution →Terminal velocity occurs when drag force equals weight, leading to $mg = kv_t$....
Read Full Step-by-Step Solution →Integrate $v$ wrt $t$: $\int_0^2 3t^2 dt = [t^3]_0^2 = 8$....
Read Full Step-by-Step Solution →Using $v^2 = u^2 + 2gh$, $v = \sqrt{2 \times 10 \times 80} = 40\,m/s$...
Read Full Step-by-Step Solution →Area under v-t graph gives displacement in the time interval....
Read Full Step-by-Step Solution →