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/s² ) Speed = 54 km/h Braking distance a =

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 The speeds are converted from km/h to m/s : u_1 = 54 km/h = 54 × (5)/(18) = 15 m/s u_2 = 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 = 6.0 + 33.33 = 39.33 m 39 m For Car B and Driver X ( u = 15 m/s , a = 7.5 m/s^2 , t_r = 0.20 s ): e = (15^2)/(2 × 7.5) = 15 m f = 15 × 0.20 + 15 = 3.0 + 15 = 18 m For Car B and Driver Y ( u = 20 m/s , a = 7.5 m/s^2 , t_r = 0.30 s ): g = (20^2)/(2 × 7.5) = 26.67 m 27 m h = 20 × 0.30 + 26.67 = 6.0 + 26.67 = 32.67 m 33 m

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Rest and Motion Kinematics

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Rest and Motion Kinematics

Determine the missing entries in the following table: Car Model...

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 $= 6.0 m/s²$ )	Speed $= 54 km/h$ Braking distance $a =$ ............................ Total stopping distance $b =$ ............................	Speed $= 72 km/h$ Braking distance $c =$ ............................ Total stopping distance $d =$ ............................
$B$ (deceleration on hard braking $= 7.5 m/s²$ )	Speed $= 54 km/h$ Braking distance $e =$ ............................ Total stopping distance $f =$ ............................	Speed $= 72 km/h$ Braking distance $g =$ ............................ Total stopping distance $h =$ ............................

Options

A(a) $18 \text{ m}$ , (b) $22 \text{ m}$ , (c) $32 \text{ m}$ , (d) $39 \text{ m}$ , (e) $15 \text{ m}$ , (f) $19 \text{ m}$ , (g) $27 \text{ m}$ , (h) $32 \text{ m}$
B(a) $19 \text{ m}$ , (b) $23 \text{ m}$ , (c) $33 \text{ m}$ , (d) $40 \text{ m}$ , (e) $14 \text{ m}$ , (f) $18 \text{ m}$ , (g) $26 \text{ m}$ , (h) $33 \text{ m}$
C(a) $20 \text{ m}$ , (b) $21 \text{ m}$ , (c) $34 \text{ m}$ , (d) $38 \text{ m}$ , (e) $16 \text{ m}$ , (f) $17 \text{ m}$ , (g) $28 \text{ m}$ , (h) $34 \text{ m}$
D(a) $19 \text{ m}$ , (b) $22 \text{ m}$ , (c) $33 \text{ m}$ , (d) $39 \text{ m}$ , (e) $15 \text{ m}$ , (f) $18 \text{ m}$ , (g) $27 \text{ m}$ , (h) $33 \text{ m}$

Correct answer

D. (a) $19 \text{ m}$ , (b) $22 \text{ m}$ , (c) $33 \text{ m}$ , (d) $39 \text{ m}$ , (e) $15 \text{ m}$ , (f) $18 \text{ m}$ , (g) $27 \text{ m}$ , (h) $33 \text{ m}$

Step-by-step solution

The speeds are converted from

\text{km/h}

\text{m/s}

u_1 = 54 \text{ km/h} = 54 \times \dfrac{5}{18} = 15 \text{ m/s}

u_2 = 72 \text{ km/h} = 72 \times \dfrac{5}{18} = 20 \text{ m/s}

The braking distance

S

is given by

S = \dfrac{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 \text{ m/s}

a = 6.0 \text{ m/s}^2

t_r = 0.20 \text{ s}

a = \dfrac{15^2}{2 \times 6.0} = 18.75 \text{ m} \approx 19 \text{ m}

b = 15 \times 0.20 + 18.75 = 3.0 + 18.75 = 21.75 \text{ m} \approx 22 \text{ m}

For Car

A

and Driver

Y

(

u = 20 \text{ m/s}

a = 6.0 \text{ m/s}^2

t_r = 0.30 \text{ s}

c = \dfrac{20^2}{2 \times 6.0} = 33.33 \text{ m} \approx 33 \text{ m}

d = 20 \times 0.30 + 33.33 = 6.0 + 33.33 = 39.33 \text{ m} \approx 39 \text{ m}

For Car

B

and Driver

X

(

u = 15 \text{ m/s}

a = 7.5 \text{ m/s}^2

t_r = 0.20 \text{ s}

e = \dfrac{15^2}{2 \times 7.5} = 15 \text{ m}

f = 15 \times 0.20 + 15 = 3.0 + 15 = 18 \text{ m}

For Car

B

and Driver

Y

(

u = 20 \text{ m/s}

a = 7.5 \text{ m/s}^2

t_r = 0.30 \text{ s}

g = \dfrac{20^2}{2 \times 7.5} = 26.67 \text{ m} \approx 27 \text{ m}

h = 20 \times 0.30 + 26.67 = 6.0 + 26.67 = 32.67 \text{ m} \approx 33 \text{ m}

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