• Home
• Class 9
• Science
• Motion
• NCERT Solution

A driver of a car travelling at 52 km/ h applies the brakes and accelerates 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?

ANSWER:

For car A:

Initial speed of the car (A) = 52 km/h × 5/18 = 14.4 m/s

From equation of motion 1^st v = u + at           

a = (v-u) / t = (0 – 14.44 ) / 5 = -2.89 m s^–2

Putting these value in equation of motion 1^st v = u + at   

t =0      v = u + at = 14.4 + (-2.89) × 0 = 14.4 m/s

t =1      v = u + at = 14.4 + (-2.89) × 1 = 11.55 m/s

t =2      v = u + at = 14.4 + (-2.89) × 2 = 8.66 m/s

t =3      v = u + at = 14.4 + (-2.89) × 3 = 5.77 m/s

t =4      v = u + at = 14.4 + (-2.89) × 4 = 2.88 m/s

t =5      v = u + at = 14.4 + (-2.89) × 5 = 0 m/s

Using these value plot graph between speed-time for car B.

From graph: Distance covered by the first car (A) before coming to rest  = Area of triangle AOB  = ½ × AO × BO  = ½ × 14.44 m/s × 5 s  = 36.1 m 

For car B:

Initial speed of the car (A) = 3 km/h × 5/18 = 0.833 m/s

From equation of motion 1^st v = u + at           

a = (v-u) / t = (0 – 0.833 ) / 10 = – 0.083 m s^–2

Putting these value in equation of motion 1^st v = u + at   

t =0      v = u + at = 0.833 + (-0.083) × 0 = 0.833 m/s

t =1      v = u + at = 0.833 + (-0.083) × 1 = 0.75 m/s

t =2      v = u + at = 0.833 + (-0.083) × 2 = 0.66 m/s

t =3      v = u + at = 0.833 + (-0.083) × 3 = 0.58 m/s

t =4      v = u + at = 0.833 + (-0.083) × 4 = 0.50 m/s

t =5      v = u + at = 0.833 + (-0.083) × 5 = 0.416 m/s

t =6      v = u + at = 0.833 + (-0.083) × 6 = 0.333 m/s

t =7      v = u + at = 0.833 + (-0.083) × 7 = 0.250 m/s

t =8      v = u + at = 0.833 + (-0.083) × 8 = 0.160 m/s

t =9      v = u + at = 0.833 + (-0.083) × 9 = 0.083 m/s

t =10      v = u + at = 0.833 + (-0.083) × 10 = 0 m/s

Using these value plot graph between speed-time for car B.

From graph: Distance covered by the second car (B) before coming to rest  = Area of triangle POQ  = ½ × PO × QO  = ½ × 0.833m/s × 10 s  = 47.2 m 

Thus, the second car (B) travels farther than the first car (A) after they applied the brakes.

Click here 🖱️👉 Motion Class 9 Quiz-1

Click here 🖱️👉 Motion Graphs Class 9 Quiz-2

Click here 🖱️👉 Motion NCERT Questions Solution

Click here 🖱️👉 Motion NCERT in-text Questions Solution

Click here 🖱️👉 Motion – additional practice Questions

Force and Laws of Motion QUIZ-1 🖱️👉 START QUIZ

Other NCERT exercise questions

A driver of a car travelling at 52 km/ h applies , A driver of a car travelling at 52 km/ h applies , A driver of a car travelling at 52 km/ h applies , A driver of a car travelling at 52 km/ h applies , A driver of a car travelling at 52 km/ h applies , A driver of a car travelling at 52 km/ h applies , A driver of a car travelling at 52 km/ h applies

A driver of a car travelling at 52 km/ h applies , A driver of a car travelling at 52 km/ h applies , A driver of a car travelling at 52 km/ h applies , A driver of a car travelling at 52 km/ h applies , A driver of a car travelling at 52 km/ h applies , A driver of a car travelling at 52 km/ h applies , A driver of a car travelling at 52 km/ h applies , A driver of a car travelling at 52 km/ h applies

हाइड्रोकार्बन यौगिक