Question

a bus starts from rest and attains a velocity of 46m/s after 10 seconds.find the acceleration and the distance travelled in 5 seconds and in 10 seconds.​

Answer 1

Answer:

1. Acceleration = 4.6 $m/s^{2}$

2. Distance at t = 5s = 57.5 m , at t = 10s = 230 m

Explanation:

First lets properly arrange the data given to us:

Given Data : u = 0 m/s  , v = 46m/s ,  t = 10s

To find :

1. Acceleration  
2. Distance at t = 10s and t = 5s

Now lets get our hands dirty and solve :

First lets calculate the acceleration :

⇒ Use the formula v = u + at

⇒ 46 = 0 + a (10)

⇒ a = $\frac{46}{10}$ = 4.6 $m/s^{2}$

Next calculate the distance at t = 10s

⇒ Use the formula $s = ut + \frac{1}{2} at^{2}$

⇒ $s = 0(10) + \frac{1}{2} 4.6 (10^{2} )$

⇒ s = 2.3 (100) = 230 m

Now distance at t = 5s

⇒ Use the formula $s = ut + \frac{1}{2} at^{2}$

⇒ $s = 0(10) + \frac{1}{2} 4.6 (5^{2} )$

⇒ s = 2.3 (25) = 57.5 m

And you have your answer as:

1. Acceleration = 4.6 $m/s^{2}$

2. Distance at t = 5s = 57.5 m , at t = 10s = 230 m

Hope it helps :)

(If you find any mistakes, please contact me within 10 minutes of writing this answer, if not possible, please forgive me)