Question

a bus is moving with a velocity of 30km/h. on applying the brakes it stops at a distance of 10m. calculate the time taken to stop the bus.


AND ANSWER IS 0.8seconds​

Answer 1

Answer:

the answer is

0.80 seconds

Hope it helps you,!!

Answer 2

$\dag \: \underline{\sf AnsWer :}$

Here we are given that, the bus is moving with a Initial Velocity (u) = 30 km/hr and bus stops when brakes are applied therefore, the Final Velocity (v) will be 0 m/s and distance (s) is given as 10 m. and we are asked to find the time taken to stop the bus.

• To solve this question first we have to make the units same. So, convert the initial velocity (u) from km/hr to m/s :

$:\implies \sf Initial \: Velocity = 30 \: km/hr$

$:\implies \sf Initial \: Velocity = 30 \times \dfrac{5}{18} \: m/s$

$:\implies \underline{ \boxed{ \sf Initial \: Velocity = \dfrac{150}{18} \: m/s}}$

So, we have converted the units to m/s and now we will find the acceleration (a) of the bus by using third kinematical equation of motion :

$\dashrightarrow\:\:\sf v^2 - u^2 = 2as$

$\dashrightarrow\:\:\sf (0)^2 - \bigg( \dfrac{150}{18} \bigg) ^2 = 2a(10)$

$\dashrightarrow\:\:\sf - \dfrac{22500}{324} = 20a$

$\dashrightarrow\:\:\sf - 22500= 20a \times 324$

$\dashrightarrow\:\:\sf - 22500= 6480a$

$\dashrightarrow\:\:\sf a = \dfrac{ - 22500}{6480}$

$\dashrightarrow\:\:\sf a = \dfrac{ - 2250}{648}$

$\dashrightarrow\:\:\sf a = \dfrac{ - 2250}{648}$

$\dashrightarrow\:\: \underline{ \boxed{\sf a = -3.473 \: m/s^2}}$

• Now, by using first kinematical equation of motion we can calculate the time taken by the bus :

$\longrightarrow\:\sf v = u + at$

$\longrightarrow\:\sf 0 = \dfrac{150}{18} + ( - 3.473 )\times t$

$\longrightarrow\:\sf - 8.34 = - 3.473 \times t$

$\longrightarrow\:\sf 8.34 =3.473 \times t$

$\longrightarrow\:\sf t = \dfrac{8.34}{3.473}$

$\longrightarrow\: \underline{ \boxed{\sf t =2.3 \: second }}$