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

Question

asked May 22, 2020 106k views

1 Answer

Fuv · Answer 1 · 2020-05-26T22:08:52+0000

Answer:

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)

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