A car moving with a speed of 40 km/h can be stopped by applying brakes after at least 2 m. If the same car is moving with a speed of 80 km/h, what is the minimum stopping distance?

Question

A car moving with a speed of 40 km/h can be stopped by applying brakes after at least 2 m. If the same car is moving with a speed of 80 km/h, what is the minimum stopping distance? (A) 4 m (B) 6 m (C) 8 m (D) 2 m

Tardigrade Admin · Accepted Answer

1 st case v2 - u2 = 2as 0-((100/9))2=2 × a × 2 [40 km/h=100/9 m/s] a = - (104/81 × 4) m/s 2nd case: 0-((200/9))2=2 × (-(104/81 × 4)) × s [80 km/h=200/9 m/s] or s = 8 m.

Q. A car moving with a speed of $40 km / h$ can be stopped by applying brakes after at least $2 m$ . If the same car is moving with a speed of $80 km / h$ , what is the minimum stopping distance?

4 m

6 m

8 m

2 m

1 st case $v^{2} - u^{2} = 2 a s$
$0 - (\frac{100}{9})^{2} = 2 \times a \times 2 [40 km / h = 100/9 m / s]$
$a = - \frac{1 0 ^{4}}{81 \times 4} m / s$
2nd case : $0 - (\frac{200}{9})^{2} = 2 \times (- \frac{1 0 ^{4}}{81 \times 4}) \times s$
$[80 km / h = 200/9 m / s]$
or $s = 8 m .$

Q. A car moving with a speed of 40km/h can be stopped by applying brakes after at least 2m. If the same car is moving with a speed of 80km/h, what is the minimum stopping distance?

4 m

6 m

8 m

2 m

1 st case v2−u2=2as 0−(9100​)2=2×a×2[40km/h=100/9m/s] a=−81×4104​m/s 2nd case : 0−(9200​)2=2×(−81×4104​)×s [80km/h=200/9m/s] or s=8m.

Q. A car moving with a speed of $40 km / h$ can be stopped by applying brakes after at least $2 m$ . If the same car is moving with a speed of $80 km / h$ , what is the minimum stopping distance?

1 st case $v^{2} - u^{2} = 2 a s$
$0 - (\frac{100}{9})^{2} = 2 \times a \times 2 [40 km / h = 100/9 m / s]$
$a = - \frac{1 0 ^{4}}{81 \times 4} m / s$
2nd case : $0 - (\frac{200}{9})^{2} = 2 \times (- \frac{1 0 ^{4}}{81 \times 4}) \times s$
$[80 km / h = 200/9 m / s]$
or $s = 8 m .$