Question

A car is moving with a speed of 50m/s in 10 sec. Another car moving with a speed of 50 m/s it takes

15 sec. Calculate and find out which covers more distance and how much?

Explain for class 9th​

Answer 1

GIVEN :-

• Velocity and time of the first car is 50 m/s in 10sec

• Velocity and time of second car is 50 m/s in 15 sec

TO FIND :-

• The car which covers more distance and by how much

SOLUTION :-

Distance covered when speed and time are given by is calculated by ,

$\large {\underline{\boxed {\red{ \bf{distance = speed \times time }}}}}$

We have ,

• u₁ = 50 m/s
• t₁ = 10 sec

Let distance covered by first car be d₁

$\implies \bf \: d_1 = u_1 \times t_1 \\ \\ \implies \bf \: d_1 = 50 \: m {s}^{ - 1} \times 10 \: s \\ \\ \implies {\underline {\boxed {\pink{ \bf{d_1 = 500 \: m}}}}}$

Distance covered by first car = 500 m

We have ,

• u₂ = 50 m/s
• t₂ = 15 sec

$\implies \bf \: d_2 = u_2 \times t_2 \\ \\ \implies \bf \: d_2 = 50 \times 15 \\ \\ \implies {\underline {\boxed {\pink{\bf{ \: d_2 = 750 \: m}}}}}$

Distance covered by second car = 750 m

$\implies \bf \: d_2 > d_1$

Difference in distance covered by two cars is ,

$\bf \: =d_2 -d_1 = 750 \: m - 500 \: m = 150m$

∴ The distance covered by the car moving with speed 50 m/s in 15 sec is more than the distance covered by the car moving with 50 m/s in 10 sec by 150 m

Answer 2

Answer:

GIVEN :-

Velocity and time of the first car is 50 m/s in 10sec

Velocity and time of second car is 50 m/s in 15 sec

TO FIND :-

The car which covers more distance and by how much

SOLUTION :-

Distance covered when speed and time are given by is calculated by ,

\large {\underline{\boxed {\red{ \bf{distance = speed \times time }}}}}

distance=speed×time

We have ,

u₁ = 50 m/s

t₁ = 10 sec

Let distance covered by first car be d₁

\begin{gathered} \implies \bf \: d_1 = u_1 \times t_1 \\ \\ \implies \bf \: d_1 = 50 \: m {s}^{ - 1} \times 10 \: s \\ \\ \implies {\underline {\boxed {\pink{ \bf{d_1 = 500 \: m}}}}}\end{gathered}

⟹d

1

=u

1

×t

1

⟹d

1

=50ms

−1

×10s

⟹

d

1

=500m

Distance covered by first car = 500 m

We have ,

u₂ = 50 m/s

t₂ = 15 sec

\begin{gathered} \implies \bf \: d_2 = u_2 \times t_2 \\ \\ \implies \bf \: d_2 = 50 \times 15 \\ \\ \implies {\underline {\boxed {\pink{\bf{ \: d_2 = 750 \: m}}}}}\end{gathered}

⟹d

2

=u

2

×t

2

⟹d

2

=50×15

⟹

d

2

=750m

Distance covered by second car = 750 m

\implies \bf \: d_2 > d_1⟹d

2

>d

1

Difference in distance covered by two cars is ,

\bf \: =d_2 -d_1 = 750 \: m - 500 \: m = 150m=d

2

−d

1

=750m−500m=150m

∴ The distance covered by the car moving with speed 50 m/s in 15 sec is more than the distance covered by the car moving with 50 m/s in 10 sec by 150 m