Question

a car initially at rest picks up a velocity of 72 km/h^-1 over a distance of 25m. Calculate acceleration of the car and time in which it picks up above velocity.​

Answer 1

Answer:

Correct Question:-

A car initially at rest picks up a velocity of $72 \: \frac{km}{h {}^{ - 1} }$ over a distance of 25 m. Calculate acceleration of the car and time in which it picks up above velocity.

Given :-

Initial velocity of car (u) = 0

Final velocity of car (v) = 72 km h$^{–1}$

Total distance covered by car (s) = 25 m.

To find:-

• Acceleration of car
• time in which it picks up above velocity.

Solution:-

Using Newton's third law of motion

$v {}^{2} - u {}^{2} = 2as$

Substitute the known values

$\Longrightarrow \: (20) {}^{2} - (0) {}^{2} = 2 \times a \times 25 \\ \\ \: \Longrightarrow a = \frac{400}{50} ms {}^{ - 2} = 8 \: ms {}^{ - 2}.$

Thus, accleration of car = 8 ms$^{–2}$

Now, time in which it picks up above velocity

Using Newton's first law of motion

$v \: = u + at$

substitute the known values

$\Longrightarrow \: 20\: = 0 + 8 + t \\ \\ \: \Longrightarrow t = \frac{20}{8} = 2.5s.$

Answer 2

Answer:

here is your answer hope it helps