Question

A bus starting from rest acquires a velocity of 15 m/s in 30 s. Calculate acceleration and distance traveled by the bus during this time. (2)

Answer 1

Answer:

acceleration=(15-0)/30=1/2m/s^2

By second equation of motion s=ut+1/2at^2=225m

pls mark brainliest

Answer 2

• initial velocity=u=0m/s
• Final velocity=v=15m/s
• Time=t=30s

We know that

$\boxed{\sf Acceleration=\dfrac{v-u}{t}}$

$\\ \sf\longmapsto Acceleration=\dfrac{15-0}{30}$

$\\ \sf\longmapsto Acceleration=\dfrac{15}{30}$

$\\ \sf\longmapsto Acceleration=0.5m/s^2$

Now using second equation of kinematics

$\boxed{\sf s=ut+\dfrac{1}{2}at^2}$

$\\ \sf\longmapsto s=0(30)+\dfrac{1}{2}\times 0.5\times (30)^2$

$\\ \sf\longmapsto s=225m$