Our top 5% students will be awarded a special scholarship to Lido.

Ncert solutions

A bus starting from rest moves with a uniform acceleration of 0.1 m s-2 for 2 minutes. Find (a)the speed acquired, (b) the distance travelled.

(a)Given, the bust starts from rest. Therefore, initial velocity (u) = 0 m/s

Acceleration (a) = 0.1 m.s-2

Time = 2 minutes = 120 s

Acceleration is given by the equation 𝑎=𝑣−𝑢𝑡

Therefore, terminal velocity (v) = (𝑎𝑡)+𝑢

(0.1 m.s-2 * 120s) + 0 m.s-1

= 12m.s-1 + 0 m.s-1

Therefore, terminal velocity (v) = 12m/s

(b)As per the third motion equation, 2𝑎𝑠=𝑣2−𝑢2

Since a = 0.1 m.s-2, v = 12 m.s-1, u = 0 m.s-1, and t = 120s, the following value for s (distance) can be obtained.

Distance, s = 𝑣2−𝑢22𝑎

= 122−022(0.1)

Therefore, s = 720m.

The speed acquired is 12m.s-1 and the total distance traveled is 720m.

(a)Given, the bust starts from rest. Therefore, initial velocity (u) = 0 m/s

Acceleration (a) = 0.1 m.s-2

Time = 2 minutes = 120 s

Acceleration is given by the equation 𝑎=𝑣−𝑢𝑡

Therefore, terminal velocity (v) = (𝑎𝑡)+𝑢

(0.1 m.s-2 * 120s) + 0 m.s-1

= 12m.s-1 + 0 m.s-1

Therefore, terminal velocity (v) = 12m/s

(b)As per the third motion equation, 2𝑎𝑠=𝑣2−𝑢2

Since a = 0.1 m.s-2, v = 12 m.s-1, u = 0 m.s-1, and t = 120s, the following value for s (distance) can be obtained.

Distance, s = 𝑣2−𝑢22𝑎

= 122−022(0.1)

Therefore, s = 720m.

The speed acquired is 12m.s-1 and the total distance traveled is 720m.

Lido

Courses

Quick Links

Terms & Policies

Terms & Policies

2021 © Quality Tutorials Pvt Ltd All rights reserved