Question

A bus starts from rest with a constant accel-
eration of 5 m/s. At the same time a car trav-
elling with a constant velocity 50 m/s over
takes and passes the bus. How fast is the bus
travelling when they are side by side?​

Answer 1

$\boxed{ \textsf{The bus takes 100 m /s</p><p> }}$

$\huge{{\huge{\boxed{\huge{\red\star\mathfrak\purple{\large{\underline{\underline{ Answer! }}}}}}}}}$

u= 0 m/s

a= 5 m/s

v= 50 m/s

t= ?

By, using the formula, we get:-

$\bold{v= u+at }$

$\bold{v= 0+5t }$

$\bold{v= 5t}$

Now,

$\bold{s= ut +\frac{1}{2}at^2}$

$\bold {\frac{5}{t^2}}$ -(1)

$\bold{v= \frac{s}{t}}$

= 50t -(2)

$\textsf{From equation -(1) and -(2)}$

$\bold{50t=\frac{5}{2t^2}}$

t= 20s

Now to find the speed of bus, we need to use the formula,

v= u+at

= 5× 20

= 100 m/s

Answer 2

SOLUTION

Let t = Time to overtake

1st case

Distance travelled by car= distance travelled by bus

$= > (ut + \frac{1}{2} at {}^{2} )car = ( ut + \frac{1}{2} at {}^{2} )bus \\ \\ = > (0)t + \frac{1}{2} (5) {t}^{2} car = (50)t + \frac{1}{2} (0) {t}^{2} bus \\ \\ = > 2.5 {t}^{2} = 50t \\ = > 2.5t = 50 \\ = > t = 20seconds$

Hence,

distance = ut

=) 50× 20

=) 1000m

2nd case

Velocity of bus:

=) v= at

=) 5× 20

=) v= 100m/s

hope it helps ☺️