Concepts Of Physics MCQ Edition [Volume 1]PhysicsRest and Motion Kinematics
Determine the missing entries in the following table: Car Model Driver X Reaction time 0.20 s Driver Y Reaction time 0.30 s A (deceleration on hard braking = 6.0 m/s ^2 ) Speed = 54 km/h Braking distance a = ............................ Total stopping distance b = ............................ Speed = 72 km/h Braking distance c = ............................ Total stopping distance d = ............................ B (
Options
- A(a) 18 m , (b) 22 m , (c) 32 m , (d) 39 m , (e) 15 m , (f) 19 m , (g) 27 m , (h) 32 m
- B(a) 19 m , (b) 23 m , (c) 33 m , (d) 40 m , (e) 14 m , (f) 18 m , (g) 26 m , (h) 33 m
- C(a) 20 m , (b) 21 m , (c) 34 m , (d) 38 m , (e) 16 m , (f) 17 m , (g) 28 m , (h) 34 m
- D(a) 19 m , (b) 22 m , (c) 33 m , (d) 39 m , (e) 15 m , (f) 18 m , (g) 27 m , (h) 33 m
Correct answer
D. (a) 19 m , (b) 22 m , (c) 33 m , (d) 39 m , (e) 15 m , (f) 18 m , (g) 27 m , (h) 33 m
Step-by-step solution
The speeds are converted from km/h to m/s : u₁ = 54 km/h = 54 5 18 = 15 m/s u₂ = 72 km/h = 72 5 18 = 20 m/s The braking distance S is given by S = u^2 2a and the total stopping distance D is given by D = u t_r + S , where t_r is the reaction time. For Car A and Driver X ( u = 15 m/s , a = 6.0 m/s ^2 , t_r = 0.20 s ): a = 15^2 2 6.0 = 18.75 m 19 m b = 15 0.20 + 18.75 = 3.0 + 18.75 = 21.75 m 22 m For Car A and Driver Y ( u = 20 m/s , a = 6.0 m/s ^2 , t_r = 0.30 s ): c = 20^2 2 6.0 = 33.33 m 33 m d = 20 0.30 + 33.33 =