Question

5.The brakes applied to a car produce an acceleration of 10 m s-2 in the opposite direction to the motion. If the car takes 5 s to stop after the application of brakes, calculate the distance it travels during this time.

Answer 1

Answer:

$\boxed{\mathfrak{Distance \ travelled = 125 \ m}}$

Given:

Accelration (a) = -10 m/s²

Final velocity (v) = 0 m/s

Time taken (t) = 5 s

To Find:

Distance travelled by car (s)

Explanation:

From equation of motion we have:

$\boxed{ \bold{s = vt - \frac{1}{2} a {t}^{2} }}$

By substituting value of v, t & a in the equation we get:

$\sf \implies s = 0 \times 5 - \frac{1}{2} \times(- 10) \times {(5)}^{2} \\ \\ \sf \implies s = 0 -(-5) \times {5}^{2} \\ \\ \sf \implies s = 5 \times 25 \\ \\ \sf \implies s = 125 \: m$

$\therefore$

Distance travelled by car (s) = 125 m

Answer 2

ANSWER

Given:

Accerleration: a=−10m/s²

Time t=5 s

Final velocity, v=0 m/s

v=u+at

0=u−10×5

u=50 m/s

s=ut+ 1/2at²

s = 50 × 5 + 1/2 (-10)×5²

s = 250 - 125

s=125 m