A particle has initial velocity `4i + 4j ms^(-1)` and an acceleration `-0.4i ms^(-2)`, at what time will its speed bbe `5ms^(-1)`?

Options

A. 2.5 s
B. 17.5 s
C. s
D. 8.5 s

Answer

A, B

Explanation

a,b. Since acceleration is in x-direction only, velocity in
y-direction will not change.
When speed `= 5 ms^(-1)`
`5^(2) = V_(x)^(2) + V_(y)^(2) = V_(x)^(2)+(4)^2`
`rArr V_x = +-3ms^(-1)`
`:. V_x = u_x + a_xt rArr t = (V_x-u_x)/(a_x)`
or `t_1 = (3-4)/(-0.4) = 2.5s`
and `t_2 = (3+4)/0.4 = 17.5s`.