Question

Brakes are applied to a moving truck to produce retardation of
5 m/s^2. If the time taken between application of the brakes and the
truck to stop is 2 seconds, calculate the distance travelled by the
truck during this time.

Answer 1

Answer:

Hey there!

Explanation:

final velocity=0

acceleration= -5m/s^2

time=2 sec

using first equation of motion,

v=u+at

0=u+(-5)(2)

10=u.

using second equation of motion,

s=ut+1/2 at^2

s=10*2+1/2*-5*2^2

s=20+1/2* -20

s=20+(-10)

s=10m

Answer 2

Given :

➔ Retardation of truck = 5m/s²

➔ Time taken to stop = 2s

To Find :

◕ Distance covered by truck before it is brought to rest.

SoluTion :

⇒ Applied brake produces retardation in the motion.

⇒ Distance covered by vehicle before it is brought to rest is called as stopping distance.

✴ Initial velocity of truck :

$\longrightarrow\tt\:v=u+at\\ \\ \longrightarrow\tt\:0=u+(-5)(2)\\ \\ \longrightarrow\tt\:u-10=0\\ \\ \longrightarrow\boxed{\bf{u=10\:ms^{-1}}}$

✴ Stopping distance :

$\implies\tt\:v^2-u^2=2as\\ \\ \implies\:(0)^2-(10)^2=2(-5)s\\ \\ \implies\tt\:s=\dfrac{100}{10}\\ \\ \implies\underline{\boxed{\bf{s=10\:m}}}$