Question

a car starts from start to rest and reaches a speed of 40 m/s in 10 s. if acceleration is constant, how fast wil it be moving in 5 s

Answer 1

Car fastly moved in 5s is

`{\bf V_(3)} = {\bf 60}(\bf m)/(\bf s)`

Explanation:

Given that a car starts from start to rest

so
`V_(1) = 0` and
`T_(1)=0`

and given car reaches a speed of 40 m/s in 10 s.

ie,
`V_(2) =40` and
`T_(2) = 10`

Use the formula

`The rate of change of velocities=acceleration* the rate of change of times`

ie,
`V_(2) - V_(1) = a(T_(2)-T_(1))`

`V_(1) = 0` ,
`T_(1) = 0` and a = constant

So,
`a = \frac {V_(2)}{T_(2)}`

`a = \frac {40}{10}`

`a = 4 \frac {m}{s}^(2)`

Similarly,

`T_(3) = 15s` (5 seconds later) and
`V_(1) = 0` ,
`T_(1) = 0`

`V_(3) - V_(1) = a* (T_(3) -T_(1))`

Then
`V_(3)-0= 4* (15-0)`

`V_(3) = 4* (15)`

`V_(3) = 60\frac {m}{s}`