Question

An object has a uniform acceleration of 3m/s^2. At a certain time its velocity is 10 m/s. What was its velocity 2 seconds earlier?

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Velocity\:2\:sec\:earlier=4\:m/s}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Acceleration(a) = - 3 { \: m/s}^{2} \\ \\ \tt: \implies Initial \: velocity(u) = 10 \: m/s \\ \\ \tt: \implies Time = 2 \: sec \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Final \: velocity(v) = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies v = u + at \\ \\ \tt: \implies v = 10 + ( - 3) \times 2 \\ \\ \tt: \implies v = 10 - 6 \\ \\ \green{\tt: \implies v = 4 \: m/s} \\ \\ \green{\tt \therefore Velocity \: before \: 2 \: second \: is \: 4 \: m/s}\\\\ \blue{\bold{Some\:related\:formula}}\\ \orange{\tt\circ\:s=ut+\frac{1}{2}at^{2}}\\\\\orange{\tt\circ\:v^{2}=u^{2}+2as}$

Answer 2

let the time at which it's velocity is 10m/s be t

in this case

Final velocity (V)=10m/s

time=t(let)

a=3

initial velocity (u)=0

since it is uniformly acelrateed motion,so,we can use the first eqaution

V=u+at

10=0+3t

t=3.3(approx)

now,we have to find it's velocity 2 sec earlier from this time t=3.3

in this case

T=3.3-2=1.3

u=0

a=3

V=0+3×1.3

V=3.9m/s or 4(approx)

that will be the velocity,of the object before two sec of the time in which it's velocity become 10m/s