Question

Your school bus accelerate uniformly from 5m/s to 10m/s in 10s. Calculate the acceleration produced and distance covered by the bus in that time.

Answer 1

Answer:

Initial velocity of the bus  (u)= 5 m/s

Final velocity of the bus v) = 10 m/s

Time is taken for a change in velocity (t)= 5 s

Therefore acceleration (a)= [10 m/s - 5 m/s]/ 5 s = 5 m/s /5s = 1 m/s²

Distance covered during these 5 seconds, is found using the relation

s = u t + ½ a t² = 5 m/s× 5 s + ½ 1 m/s² × 5² s²=25 m+ 12.5 m = 37.5 m

So the bus covers 37.5 m during the period its velocity changes from 5 m/s to 10 m/s.

Answer 2

$\bigstar\huge \mathtt{\underline{ Solution : } }$

As per the given data ,

• Initial velocity of the bus = 5 m/s
• Final velocity of the bus = 10 m /s
• Time taken = 10 s

We need to find the acceleration and the distance covered by the bus in that time

As the bus is moving with uniform acceleration throughout it's motion we can use kinematic equation in order to solve such questions

Let's find the acceleration of the car by using first equation of motion ,

➽ v = u + at

here ,

• v = final velocity
• u = initial velocity
• a = acceleration
• t = time

Now let's substitute the given values in the above equation ,

➝ 10 = 5 + a x 10

➝ 10 - 5 = 10 a

➝ 10 a = 5

➝ a = 0.5 m / s²

The acceleration of the bus is 0.5 m/s²

Now let's find the distance covered by the bus in 10 s in order to so that simply use second equation of motion

➽ s = ut + 1/2 at²

here ,

• s = distance
• u = initial velocity
• a = acceleration
• t = time taken

➝ s = 5 x 10 + 0.5x 100 / 2

➝ s = 50 + 100 / 4

➝ s = 50 + 25

➝ s = 75 m

The distance covered by the bus is 75 m