5. A driver of a car travelling ai 52 km h applies the brakes and
celerates uniformly in the opposite direction. The car stops
in 5 s Another driver going at 3 km h' in another car applies
his brakes slowly and stops in 10 s. On the same graph paper,
plot the speed versus time graphs for the two cars. Which of
the two cars travelled farther after the brakes were applied?
Answers
GIVEn
A driver of a car travelling ai 52 km h applies the brakes and acccelerates uniformly in the opposite direction. The car stops
in 5 s Another driver going at 3 km h' in another car applies his brakes slowly and stops in 10 s.
TO FINd
Which of the two cars travelled farther after the brakes were applied?
SOLUTIOn
★ In first case ★
- u = 52km/h
- v = 0
- t = 5s
- a = ?
- s = ?
→ u = 52 km/h
→ u = 52 × 5/18
→ u = 14.4 m/s
**Apply formula**
→ v = u + at
→ 0 = 14.4 + a × 5
→ 14.4 + 5a = 0
→ 5a = -14.4
→ a = -14.4/5 = - 2.8m/s²
Now the value of " s "
**Again apply formula**
→ s = ut + ½ at²
→ s = 14.4 × 5 + ½ × (-2.8) × (5)²
→ s = 72 - 35
→ s = 37m
_____________________
★ In second case ★
- u = 3km/h
- v = 0
- t = 10s
- a = ?
- s = ?
→ u = 3km/h
→ u = 3 × 5/18
→ u = 15/18
→ u = 0.8m/s
**Apply this formula**
→ v = u + at
→ 0 = 0.8 + a × 10
→ 0 = 0.8 + 10a
→ 10a = -0.8
→ a = -0.8/10 =- 0.08 m/s²
Now , the value of " s "
**Again apply formula**
→ s = ut + ½ at²
→ s = 0.8 × 10 + ½ × (-0.08 )× (10)²
→ s = 8 - 4
→ s = 4
Hence, first driver of car travelled father after applied break
Answer:
Car 1 Travelled farther when breaks were applied.
Explanation:
Car 1
Distance travelled
= Area under v-t
=1/2 *b*h
1/2 * 5* 130/9
= 36.11m
Car 2
Distance Travelled
= Area under v-t
= 1/2 * b*h
= 1/2 *10* 5/6
=4.166m
