Question

A Scooter starting from rest moves with a Constant acceleration and covers a distance of 64m in 4 sec. calculate
a) its acceleration & final velocity
b) At what time The scooter had covered half the total distance?​

Answer 1

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

$\green{\tt{\therefore{Acceleration=8\:m/s^{2}}}}$

$\green{\tt{\therefore{Final\:velocity=32\:m/s}}}$

$\green{\tt{\therefore{Time\:half=2\sqrt{2}\:sec}}}$

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

$\green{\underline \bold{Given :}} \\ \tt: \implies Distance(s) = 64 \: m \\ \\ \tt: \implies Time(t) = 4 \: sec \\ \\ \red{\underline \bold{To \:Find :}} \\ \tt: \implies Acceleration(a) =? \\ \\ \tt: \implies Final \: velocity(v) =? \\ \\ \tt: \implies Time \: taken \: to \: cover \: half \: distance(t_{1}) = ?$

• According to given question :

$\tt \circ \: Initial \: velocity = 0 \: m/s \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s = ut + \frac{1}{2} a {t}^{2} \\ \\ \tt: \implies 64 = 0 \times 4 + \frac{1}{2} a \times {4}^{2} \\ \\ \tt: \implies 64 \times 2 = 16a \\ \\ \tt: \implies a = \frac{128}{16} \\ \\ \green{\tt: \implies a = 8 { \: m/s}^{2} } \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies v = u + at \\ \\ \tt: \implies v = 0 + 8 \times 4 \\ \\ \green{\tt: \implies v = 32 \: m/s} \\ \\ \tt \circ \: Distance = 32 \: m \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s_{1} = u t_{1}+ \frac{1}{2} a { t_{1}}^{2} \\ \\ \tt: \implies 32 = 0 \times t_{1} + \frac{1}{2} \times 8 \times { t_{1} }^{2} \\ \\ \tt: \implies \frac{32 \times 2}{8} = { t_{1} }^{2} \\ \\ \tt: \implies t_{1} = \sqrt{8} \\ \\ \green{\tt: \implies t_{1} = 2 \sqrt{2} \: sec}$

Answer 2

Explanation:

Acceleration = 8m/s^2.

Final Velocity= 32m/s.

Therefore, Time= 2√2 sec.