Question

A car is going at 60 km per hour is caused to stop in 3 seconds its average deceleration in kilometre per hour second is​

Answer 1

Initial Velocity(u): 60kmph
Time(t):3 sec
Final Velocity(v): 0kmph (as the car stops after applying brakes)

Deceleration= v-u/t
= 0-60/3
-20 Kilometres per hour square (-20Km/h^2)
Now it is said deceleration so it will be 20km/h^2 (when asked deceleration or retardation, remove the minus, negative sign as the term means that.)

Thus, Deceleration will be 20km/h^2 (-20km/h^2 in case of acceleration)

Answer 2

Answer:

a =  $\frac{50}{9}$ $ms^{-2}$

a = $72000$ $Kmh^{-2}$

a = $20$ $Km(hour second)^{-1}$

Explanation:

initial velocity , u = 60 km/h = $\frac{60 * 1000}{3600}$ = $\frac{100}{6}$ m/s

final velocity, v = 0

time taken = 3s = $\frac{3}{3600}$ hour = $\frac{1}{1200}$ hour

deceleration , a = ?

we can use the formula

v = u + at

1. In $ms^{-2}$ :

0 = $\frac{100}{6}$ + $3a$

a = $\frac{ - 100}{6 * 3}$ =  $\frac{ - 100}{18}$ = $\frac{ - 50}{9}$ $ms^{-2}$

2. In $Kmh^{-2}$ :

0 = $60$ + $\frac{1a}{1200}$

a = $(- 60 * 1200)$ =  = $-72000$ $Kmh^{-2}$

3. In $Km(hour second)^{-1}$ :

0 = $60$ + $3a$

a = $\frac{-60}{3}$ =  = $-20$ $Km(hour second)^{-1}$