Question

A bas moving with velocity of 60 kmhe
brought to rest in aoo by applying
beakes. Find the acceleeation.​

Answer 1

U= 60km/hr × 5/18 = 50/3 m/sec

U= 60km/hr × 5/18 = 50/3 m/secv= 0

U= 60km/hr × 5/18 = 50/3 m/secv= 0t= 20s

U= 60km/hr × 5/18 = 50/3 m/secv= 0t= 20sa= ?

U= 60km/hr × 5/18 = 50/3 m/secv= 0t= 20sa= ?we know that-

U= 60km/hr × 5/18 = 50/3 m/secv= 0t= 20sa= ?we know that-a = v-u/t

U= 60km/hr × 5/18 = 50/3 m/secv= 0t= 20sa= ?we know that-a = v-u/ta = -50/3 /20

U= 60km/hr × 5/18 = 50/3 m/secv= 0t= 20sa= ?we know that-a = v-u/ta = -50/3 /20a= -50/60

U= 60km/hr × 5/18 = 50/3 m/secv= 0t= 20sa= ?we know that-a = v-u/ta = -50/3 /20a= -50/60∴ a = -0.83m/sec²

U= 60km/hr × 5/18 = 50/3 m/secv= 0t= 20sa= ?we know that-a = v-u/ta = -50/3 /20a= -50/60∴ a = -0.83m/sec²negative sign shows that it is a retarded motion.

U= 60km/hr × 5/18 = 50/3 m/secv= 0t= 20sa= ?we know that-a = v-u/ta = -50/3 /20a= -50/60∴ a = -0.83m/sec²negative sign shows that it is a retarded motion.so its acceleration is 0.83 m/sec²