Question

IF BREAKS ARE APPLIED TO A CAR PRODUCE AN ACCELERATION OF 10M/S IN THE OPPOSITE DIRECTION OF THE MOTION . IF IT TAKES 5 SCEONDS TO STOP AFTER THE APPLICATION OF BREAKS. FIND THE DISTANCE TRAVELLED BY THE CAR BEFORE COMING TO REST?

Answer 1

a=-10m/s

v=0

t=5s

u= ?

according to first equation

v = u+at

0=u+-25

u = 25 m/s

According to third equation

v2 -u2 = 2as

0 - 625 = 2*-10*s

-625/-20 = S

S = Distance = 31.25

$....$

Answer 2

Answer:

Explanation:

This can be solved by using first and second equation of motion.

a = -10m/s^2

t = 5 seconds

v = 0 m/s

So we can take out u by using first equation of motion i.e.

v = u + at

0 = u + -10× 5

0 = u - 50

50 = u

u = 50 m/s

Now by second equation of motion we can find the distance travelled .

Which you can see in the attachment.