Question

a car starting from rest on road accelerator in a straight line at a constant rate of 7m s-2 for 10.0s. how far does the car travel during this time





answer le fast plz....​

Answer 1

Provided that:

• Initial velocity = 0 mps
• Acceleration = 7 mps sq.
• Time = 10 seconds

Don't be confused! Initial velocity cames as zero because the car starts from rest.

To calculate:

• Distance travelled

Solution:

• Distance travelled = 350 m

Using concept:

• Second equation of motion

Using formula:

• ${\small{\underline{\boxed{\pmb{\sf{s \: = ut \: + \dfrac{1}{2} \: at^2}}}}}}$

Where, s denotes displacement or distance or height, u denotes initial velocity, t denotes time taken and a denotes acceleration.

Required solution:

$:\implies \sf s \: = ut \: + \dfrac{1}{2} \: at^2 \\ \\ :\implies \sf s \: = 0(10) + \dfrac{1}{2} \times 7(10)^{2} \\ \\ :\implies \sf s \: = 0(10) + \dfrac{1}{2} \times 7(100) \\ \\ :\implies \sf s \: = 0 + \dfrac{1}{2} \times 700 \\ \\ :\implies \sf s \: = \dfrac{1}{2} \times 700 \\ \\ :\implies \sf s \: = 1 \times 350 \\ \\ :\implies \sf s \: = 350 \: m \\ \\ :\implies \sf Distance \: travelled \: = 350 \: m$

• It can go 350 m!

Additional information:

• The three equations of motion:

$\begin{gathered}\boxed{\begin{array}{c}\\ {\pmb{\sf{Three \: equation \: of \: motion}}} \\ \\ \sf \star \: v \: = u \: + at \\ \\ \sf \star \: s \: = ut + \: \dfrac{1}{2} \: at^2 \\ \\ \sf \star \: v^2 - u^2 \: = 2as\end{array}}\end{gathered}$

• Something about distance and displacement:

• Distance = It is the length of actual path covered by a moving object in a given time interval.

• Displacement = Shortest distance covered by a body in a definite direction is called displacement.

→ Displacement may be positive, negative or zero whereas distance is always positive.

→ Distance is a scaler quantity whereas displacement is a vector quantity both having the same unit.

→ The SI unit of distance and displacement is m/s.

→ In general magnitude of displacement ≤ distance.

Answer 2

Answer:

350 m

Explanation:

As per the provided information in the given question, we have :

• Initial velocity (u) = 0 m/s
• Acceleration (a) = 7 m/s²
• Time (t) = 10 s

We've been asked to calculate distance travelled.

By using the second equation of motion,

⇒ s = ut + ½at²

• s denotes distance
• u denotes initial velocity
• t denotes time
• a denotes acceleration

⇒ s = 0(10) + ½ × 7 × (10)²

⇒ s = ½ × 7 × 100 m

⇒ s = 1 × 7 × 50 m

⇒ s = 350 m

Therefore, the distance travelled is 350 m.

$\rule{200}2$

Learn More :

First equation of motion :

• v = u + at

Second equation of motion :

• s = ut + ½at²

Third equation of motion :

• v² – u² = 2as