Question

A car is moving with a uniform velocity of 72 km/h which is then brought to rest by applying brakes after 10s. Find the deceleration of the car?

Answer 1

DATA :..
Vi=72km/hr =20m/s
t=10s
a=?
SOLUTION:..
Vf=Vi+at
a=Vf-Vi/t
i.e ..Vf=0
a= 20/10=
2m/s^2

Answer 2

Answer:

2 m/s

Explanation:

deceleration= $-\frac{v-u}{t}$

convert 72 km/h to m/s as the time is in m/s

we get $\frac{72}{1}$ x $\frac{5}{18}$

we get 4 x 5 = 20

initial velocity = 20 m/s

now lets solve

-$-\frac{0-20}{10}\\\\= -\frac{-20}{10}\\= -(-2)\\= 2m/s$

deceleration = 2m/s