Question

4. A car is moving with a speed of 10 m/s. When the brakes are applied, the car has a
constant acceleration of 4 m/s. What is the distance travelled before the car comes to
rest? Also calculate the time taken by the car to come to rest. (SCAN AND UPLOAD)​

Answer 1

Answer:

4. A car is moving with a speed of 10 m/s. When the brakes are applied, the car has a

constant acceleration of 4 m/s. What is the distance travelled before the car comes to

rest? Also calculate the time taken by the car to come to rest. (SCAN AND UPLOAD)

Explanation:

4. A car is moving with a speed of 10 m/s. When the brakes are applied, the car has a

constant acceleration of 4 m/s. What is the distance travelled before the car comes to

rest? Also calculate the time taken by the car to come to rest. (SCAN AND UPLOAD)

Answer 2

${\huge{\bigstar{\blue{\rm{GIVEN}}}}}\bigstar$

• $\sf{Initial\:Velocity = 10m/s}$

• $\sf{Retardation = 4m/s}$

• $\sf{Final\:Velocity = 0m/s}$

${\huge{\bigstar{\red{\rm{TO\:FIND}}}}}\bigstar$

• The Distance travelled and the time taken by car to comes to rest.

${\huge{\bigstar{\red{\rm{FORMULAE\:USED}}}}}\bigstar$

• ${\huge{\boxed{\sf{v^2 - u^2 = 2as}}}}$

• ${\huge{\boxed{\sf{ v = u + at}}}}$

Where,

v= Final Velocity

u = Initial Velocity

a = Accceleration

S = Distance

t = Time

${\huge{\red{\underline{\sf{Solution}}}}}$

$\implies\sf{v^2 - u^2 = 2as}$

$\implies\sf{ (0)^2 - (10)^2 = 2 * 4 * s}$

$\implies\sf{ 0 - 100 = -8s}$

$\implies\sf{ -100 = -8s}$

$\implies\sf{ S = \cancel{\dfrac{100}{8}}}$

$\implies\sf{ S = 12.5m}$

Hence, The Distance Travelled by car after being brakes are applied is 12.5m.

Therefore,

$\implies\sf{ v = u + at}$

$\implies\sf{ (0) = (10) + -4 × t}$

$\implies\sf{ -10 = -4t}$

$\implies\sf{ t = \cancel{\dfrac{10}{4}}}$

$\implies\sf{ t = 2.5s}$.

Hence, The time taken by car to stop is 2.5s.