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A particle is moving in xy-plane .At cer...

A particle is moving in xy-plane .At certain instant the components of its velocity and acceleration are as follows `u_(x) = 3m//s,u_(y)=4m//s,a_(x)=2m//s^(2)"and"a_(y)=1m//^(2).` The rate of change of speed at this moment is
`(a) 4.m//s^(2)`
`(b) 2,m//s^(2)`
`(c) sqrt3.m//s^(2)`
`(d) sqrt5 .m//s^(2)`

Text Solution

Verified by Experts

Strategy : Component of acceleration along the direction of velocity is called rate of change of speed.
Given that ` v_(x) = 3m//s , v_(y) = 4 m//s , a_(x) 2m//s^(2) , a_(y) = 1 m//s^(2)`
` vecv = v_(x) hati + v_(y)hatj = 3hati + 4 hatj , veca ,= a_(x)hati + a_(y)hatj = 2hati + 1hatj`
` vecv,veca = ( 3hati + 4hatj) ( 2hati + 1hatj) = 6+ 4 =10`
` vacos theta =10`
` a cos theta = 10/v = 10/5 =2`
Rate of change of speed = ` a cos theta = 2 m//s`
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