Question

A body with initial velocity 5m/sec moves along a straight line with constant acceleration and travels 420 meters in 30 seconds. Calculate the average speed of the body during the time period (s), the final velocity of the body (V(subscript)f) and the acceleration on the body (a).
S=
V(sub)f=
a=
​

Answer 1

Answer:

initial velocity, u = 5m/s

distance, s = 420m

time, t = 30s

1) Avg speed = 420/30 = 14m/s

2) s = ut + 1/2 a$t^{2}$ ;  420 = 5(30) + 1/2*(900)*a = 150 + 450a

                               450a = 270,   a = 270/450 = 6/10 =0.6m/$s^{2}$

3) v = u +at ;  v = 5 + 0.6*30 = 5 + 18 = 23m/s

Answer 2

$\maltese\:\underline{\sf AnsWer :}\:\maltese$

A body with Initial Velocity (u) 5 m/s moves along a straight line with constant acceleration and travels distance (s) = 420 meters in time Interval (t) = 30 seconds.

We need to find the three things here :

1. Average speed of the body.
2. The final velocity (v)
3. Acceleration (a)

First let's find the Average Speed of the body and then by using the kinematical equation of motion we can find the final velocity (v) and Acceleration (a) of the body.

$\dag\: \pink{\underline{\tt \gray{ Average \: Speed \: (V_{av}) : }}}$

$\longrightarrow\:\:\sf Average \: Speed = \dfrac{Total \: distance \: traveled \: by \: the \: body}{Total \: time \: taken \: by \: the \: body}$

$\longrightarrow\:\:\sf Average \: Speed = \dfrac{420}{30}$

$\longrightarrow\:\:\sf Average \: Speed = \dfrac{42}{3}$

$\longrightarrow\:\: \underline{\boxed{ \textsf{\textbf {Average Speed = 14 m{s} ^{ \text{-1}} }}}}$

Now, by using second kinematical equation of motion we can find the Acceleration of the body

$\dag\: \pink{\underline{\tt \gray{ Acceleration \:of \:the\:body \: (a) : }}}$

$\dashrightarrow\:\:\sf s = ut + {}^{1} \! / {}_{2} \: at^2$

$\dashrightarrow\:\:\sf 420=5 \times 30+ {}^{1} \! / {}_{2} \times a \times {(30)}^{2}$

$\dashrightarrow\:\:\sf 420=150+ {}^{1} \! / {}_{2} \times a \times900$

$\dashrightarrow\:\:\sf 420 - 150 = {}^{900} \! / {}_{2} \times a$

$\dashrightarrow\:\:\sf 270= 450 \times a$

$\dashrightarrow\:\:\sf a = \frac{270}{450}$

$\dashrightarrow\:\:\sf a = \frac{27}{45}$

$\dashrightarrow\:\: \underline{\boxed{ \textsf{\textbf {Acceleration = 0.6 m{s} ^{ \text{-2}} }}}}$

$\dag\: \pink{\underline{\tt \gray{Final\: velocity\:of \:the\:body \: (v) : }}}$

$:\implies\sf Acceleration = \dfrac{Final \: Velocity - Initial \: Velocity}{Time \: taken}$

$:\implies\sf 0.6 = \dfrac{Final \: Velocity - 5}{30}$

$:\implies\sf 0.6 \times 30 =Final \: Velocity - 5$

$:\implies\sf 18=Final \: Velocity - 5$

$:\implies\sf Final \: Velocity = 18 + 5$

$:\implies \underline{\boxed{ \textsf{\textbf {Final Velocity = 23 m{s} ^{ \text{-1}} }}}}$