A body is moving with velocity 30 m-s towards east- After 10 seconds its velocity becomes 40 m-s towards north- The average acceleration of the body is

Question

Toppr Admin · Accepted Answer

Initial velocity-xA0- u-30-iFinal velocity-xA0-xA0- v-40-jChange in velocity-xA0- -x394-v-40-j-x2212-30-iMagnitude-xA0-xA0-xA0- -x394-v-x221A-302-402-50Average acceleration-xA0-xA0- a-x394-v-x394-t-5010-5-xA0-m-s2

A body is moving with velocity 30 m/s towards east. After 10 seconds its velocity becomes 40 m/s towards north. The average acceleration of the body is

Initial velocity $u = 30^i$
Final velocity $v = 40^j$
Change in velocity $Δ v = 40^j - 30^i$
Magnitude $| Δ v | = \sqrt{30^{2} + 40^{2}} = 50$
Average acceleration $a = \frac{| Δ v |}{Δ t} = \frac{50}{10} = 5 m / s^{2}$

A body is moving with velocity 30 m/s towards east. After 10 seconds its velocity becomes 40 m/s towards north. The average acceleration of the body is

Initial velocity u=30^iFinal velocity v=40^jChange in velocity Δv=40^j−30^iMagnitude |Δv|=√302+402=50Average acceleration a=|Δv|Δt=5010=5 m/s2

Initial velocity $u = 30^i$
Final velocity $v = 40^j$
Change in velocity $Δ v = 40^j - 30^i$
Magnitude $| Δ v | = \sqrt{30^{2} + 40^{2}} = 50$
Average acceleration $a = \frac{| Δ v |}{Δ t} = \frac{50}{10} = 5 m / s^{2}$