Question

A car moving with a uniform velocity of 100m/s is decelerated at deceleration of 2.5m/s square to stop. Calculate the time taken for the car to stop and the distance travelled by the car before it brought to rest. Please help me

Answer 1

Answer:

u = 100m/s

a = -2.5m/s^2

v = 0

t = ?

s = ?

t = v - u / a

= 0 - 100m/s / -2.5m/s^2

= 40s

s = ut + 1/2at^2

= 100m/s * 40s + 1/2 * -2.5m/s^2 * ( 40s )^2

= 4000m + 1/2 * -2.5m/s^2 * 1600s^2

= 4000m + 1/2 * -4000m

= 4000m + ( -2000m)

= 4000m - 2000m

= 2000m

hope it helps

Answer 2

The time taken for the car to stop and the distance traveled by the car before it is brought to rest are 40sec and 2km respectively.

Given:-

The initial velocity of car (u) = 100m/s

Deceleration of car (-a) = 2.5m/s²

The final velocity of car (v) = 0m/s

To Find:-

The time taken for the car to stop and the distance traveled by the car before it is brought to rest.

Solution:-

We can simply calculate the time taken for the car to stop and the distance traveled by car before it is brought to rest by using the following method.

As

The initial velocity of car (u) = 100m/s

Deceleration of car (-a) = 2.5m/s²

The final velocity of car (v) = 0m/s

Time taken (t) =?

Distance travelled by car before rest (s) =?

According to the first equation of motion,

$v = u + at$

$0 = 100 - 2.5 \times t$

$2.5 \times t = 100$

$t = \frac{100}{2.5}$

$t = \frac{1000}{25}$

$t = 40sec$

Now, By using the Third equation of motion,

${v}^{2} = {u}^{2} + 2as$

${0}^{2} = {100}^{2} - 2 \times 2.5 \times s$

$10000 = 2 \times 2.5 \times s$

$10000 = 5s$

$s = \frac{10000}{5}$

$s = 2000m$

$s = 2km$

Hence, The time taken for the car to stop and the distance traveled by the car before it is brought to rest are 40sec and 2km respectively.

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