Answer to Question #96282 in Mechanics | Relativity for Joseph

Question

A body at rest is at initial acceleration for 30s. After 30s which acceleration is reduced to 5m/s^2 for the bext 20s. The body then maintains the speed attained for 60s after which brought to rest in 20s. Draw the velocity time graph of the motion using the information given above to calculate,(a)maximum speed (b)average as the body is been brought to test (c) total distance travelled during the same 50s (d) average speed during the same interval

Answer 1

A body ,at rest,is given an initial acceleration of $8 m/sec^2$ for $30 \ sec$

Velocity after $30 sec=8 \times30=240m/sec$

For $t=30$ to $t=50 sec;a=5m/sec^2$

( a ) Velocity after $50$ seconds$=240+20\times5=340m/sec^2$ =Maximum Velocity

For $t=50 \ to\ t=110;a=0;$ velocity remains constant.

For $t=110 \ to \ t=130;$ velocity is reduced down to zero.

$v=u+at\implies 0=340+a\times 20 \implies a=-17m/sec^2$

The area of $v-t$ graph determines the total distance traveled.

The area of graph is $=\frac{1}{2}\times 30 \times 240+240 \times 20+\frac{1}{2}\times 20\times100+60\times340$ $+\frac{1}{2}\times20\times340=33200m=33.3Km$

( b ) Average velocity $=\frac{distance}{time}=\frac{33200}{130}=255.38m/sec$

(c) Distance traveled in first $50$ seconds$==\frac{1}{2}\times 30 \times 240+240 \times 20+\frac{1}{2}\times 20\times100=9400m$

(d) Average speed in first $50$ seconds$=\frac{9400}{50}=188m/sec$