Question

plzzz answer these questions
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Answer 1

Answer:

1] Unfortunately there’s no single equation to calculate this in a way similar to the SUVAT equations as far as I know, so I’ll be using logic to pick this problem apart initially.

So, given that Car A is travelling at 60 km/h and travels for an hour before Car B starts, we can use distance = speed * time to calculate he distance Car A is ahead of Car B the exact moment Car B begins moving.

speed = 60 kilometres per hour

time = 1 hour

distance = 60 * 1

distance = 60 kilometres

Hence, Car A is exactly 60 kilometres ahead of Car B when Car B begins moving.

As Car B is travelling 15 kilometres per hour faster than Car A (75 - 60 = 15), we simply have to calculate the time it will take to cover 60 kilometres at 15 kilometres per hour.  

Using the same equation as previously, except this time we’re working out time instead of distance:  

speed = 15 kilometres per hour  

distance = 60 killmetres  

60 = 15 * time

time = 60 / 15

time = 4 hours

Therefore, it will take Car B 4 hours to catch up to and overtake Car A.

5] option a

6] option c

7] option c

8

u=0

v=40m/s

t=20s

By using the formula,

v=u+at

40=0+a×20

40=20a

a=2m/s^2

9]

actually value of acceleration and deceleration are same but their sign are not same ...

acceleration have positive sign where as deceleration have negative sign..

for example the the deceleration given in the question is tw0 therefore the value of acceleration is 2

10]  

By using the equation: v = u +at

At highest point, velocity, v = 0 m/s and acceleration, a = -g = -9.8 m/s^2.

Initially the velocity, u = 49 m/s.

Therefore, 0 = 49 - 9.8t => t = (-49)/(-9.8) = 5 seconds.

i hope it helps u