Answer to Question #241591 in Mechanics | Relativity for Abioye Oluwabukola

Question

A body at rest is given an initial uniform acceleration of 8m/s^2 for 30sec after which the acceleration is increased to 5m/s^2 for the next 20sec.calculate the maximum speed attained during motion,average retardation as the body is been brought to rest,total distance travelled during the first 50sec and the average speed during the same interval

Answer 1

a) calculate the maximum speed

$u=0m/s, a=8m/s^2, t=30s$

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

$8=\frac{v-0}{30}$

$v=8\times 30=240m/s$

new acceleration =8 +5 =$13m/s^2$

for the next 20 seconds

$a=13m/s^2, t=20s, u=240m/s, v=?$

$13=\frac{v-240}{20}$

$260=v-240$

$v=500m/s(max \space speed)$

b) calculate average retardation

$u=500m/s, v=0m/s, a=?$

if we assume the average retardation took 20seconds, then

$a=\frac{0-500}{20}=-25m/s^2$

c) total distance traveled during the first 50sec

total distance $=$ distance during the first 30 seconds + distance during the 20 seconds period

$=240\times 30+ 500\times 20=7200+10000=17200m$

d) average speed $=\frac{240m/s+500m/s}{2}=370m/s$