Question

A car initially at rest starts moving with a constant acceleration of 0.5ms^-2 and travels a distance 25 cm .calculate its final velocity and time taken.​

Answer 1

• We have,

Acceleration(a) = 0.5m/s^-2

Distance travelled(d) = 25m

Initial velocity(u) = 0m/s

• We have to find,

1.Final velocity

2.Time taken

${\sf{\pink {\underline{\large{\pink{\maltese\:Final \: velocity}}}}}}$

We know that,

${\green{\boxed{\sf{v^2=u^2 + 2as \tiny \blue{(Third \: equation \: of \: motion)}}}}}$

Putting values in the formula

$\sf \rightsquigarrow \: \: {v^2 = (0^2) + 2×0.5×25}$

$\sf \rightsquigarrow \: \: {v^2 = 0 + 2 \times 0.5 \times 25}$

$\sf \rightsquigarrow \: \: {v^2 = 0 + 25}$

$\sf \rightsquigarrow \: \: {v^2 = 25}$

$\sf \rightsquigarrow \: \: {v^2-25 = 0}$

$\sf \rightsquigarrow \: \: {(v - 5) \: (v + 5) = 0}$

$\sf \rightsquigarrow \: \: {v = 5ms^{-1}}$

${\sf{\pink {\underline{\large{\pink{\maltese \: Time \: taken}}}}}}$

We know that,

${\green{\boxed{\sf{v=u + at \tiny \blue{(1st\: equation \: of \: motion)}}}}}$

Putting values in the formula

$\sf \rightsquigarrow \: \:{5 = 0 + 0.5t}$

$\sf \rightsquigarrow \: \:{0.5t = 5}$

$\sf \rightsquigarrow \: \:{t = \frac{5}{0.5} }$

$\sf \rightsquigarrow \: \:{t = \cancel\frac{50^{10}}{5_1} }$

$\sf \rightsquigarrow \: \:{t = 10s}$

Also view :

https://brainly.in/question/44895255

https://brainly.in/question/44927333

Answer 2

Hello swarish good morning hru and hope it will help you✨ have a beautiful morning