Question

Will mark brainlest helpp!!!!!!​

Answer 1

• initial velocity=u=0m/s
• Final velocity=60km/h=v

Convert to m/s

$\\ \Large\sf\longmapsto v=60\times \dfrac{5}{18}=16.6m/s$

• Time=t=5min=5×60=300s

We know

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

$\\ \Large\sf\longmapsto Acceleration=\dfrac{16.6-0}{300}$

$\\ \Large\sf\longmapsto Acceleration=\dfrac{16.6}{300}$

$\\ \Large\sf\longmapsto Acceleration=0.05m/s^2$

• Distance=s

Using second equation of kinematics

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

$\\ \Large\sf\longmapsto s=0(300)+\dfrac{1}{2}(0.05)(300)^2$

$\\ \Large\sf\longmapsto s=0.05\times 150^2$

$\\ \Large\sf\longmapsto s=0.05\times 22500$

$\\ \Large\sf\longmapsto s=1125m$

Answer 2

Answer:

I don't know if it is correct or not.