Question

JEE MAINS PHYSICS QUESTION SEPTEMBER 2020

Answer 1

Answer:

20 m

Explanation:

area under vt graph gives distance

=1/2*5*8

=20 m

hope gonna help

Answer 2

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

$\green{\tt{\therefore{Distance\:travelled=20\:m}}}$

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

$\green{\underline \bold{Given :}} \\ \tt: \implies Final \: velocity(v) = 8 \: ms \\ \\ \tt: \implies Time(t) = 5 \: sec \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Distance \: travelled \: by \: particle(s) = ?$

• According to given question :

$\tt \circ \: Initial \: velocity(u) = 0 \: m/s \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies v = u + at \\ \\ \tt: \implies 8 = 0 + a \times 5 \\ \\ \tt: \implies a = \frac{8}{5} \: m/ {s}^{2} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s = ut + \frac{1}{2} a {t}^{2} \\ \\ \tt: \implies s = 0 \times 5 + \frac{1}{2} \times \frac{8}{5} \times {5}^{2} \\ \\ \tt: \implies s = 4 \times 5 \\ \\ \green{\tt: \implies s = 20 \: m} \\ \\ \bold{Alternate \: method : } \\ \tt: \implies Distance \: trvavelled = Area \: under \: curve \\ \\ \tt: \implies Distance \: travelled = \frac{1}{2} \times base\times height \\ \\ \tt: \implies Distance \: travelled = \frac{1}{2} \times 5 \times 8 \\ \\ \green{\tt: \implies Distance \: travelled =20 \: m} \\ \\ \green{\tt \therefore Distance \: travelled \: by \: particle \: is \: 20 \: m}$