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19 votes
19 votes
A car starts from rest and travels for 5.0 s with a constant acceleration of

-1.5 m/s. What is the final velocity of the car? How far does the car travel
in this time interval?

User Amir Afianian
by
2.8k points

1 Answer

12 votes
12 votes

Step-by-step explanation:

Step-by-step explanation:

If an object having initial velocity

uExplanation:

If an object having initial velocity

u

is applied an acceleration of

a

for

t

time, its final velocity

v

is given by

v

=

u

+

a

t

and distance covered

S

is given by

S

=

u

t

+

1

2

a

t

2

.

We have

u

=

o

(as the object starts from rest),

t

=

5

sec., and acceleration

a

=

1.5

m

sec

2

.

Hence final velocity

v

=

1.5

×

5

=

7.5

m

sec

.

Distance covered

S

=

1

2

×

(

1.5

)

×

5

2

=

18.75

m

.

Minus sign indicates that acceleration is in reverse direction.

is applied an acceleration of

a

for

t

time, its final velocity

v

is given by

v

=

u

+

a

t

and distance covered

S

is given by

S

=

u

t

+

1

2

a

t

2

.

We have

u

=

o

(as the object starts from rest),

t

=

5

sec., and acceleration

a

=

1.5

m

sec

2

.

Hence final velocity

v

=

1.5

×

5

=

7.5

m

sec

.

Distance covered

S

=

1

2

×

(

1.5

)

×

If an object having initial velocity

u

is applied an acceleration of

a

for

t

time, its final velocity

v

is given by

v

=

u

+

a

t

and distance covered

S

is given by

S

=

u

t

+

1

2

a

t

2

.

We have

u

=

o

(as the object starts from rest),

t

=

5

sec., and acceleration

a

=

1.5

m

sec

2

.

Hence final velocity

v

=

1.5

×

5

=

7.5

m

sec

.

Distance covered

S

=

1

2

×

(

1.5

)

×

5

2

=

18.75

m

.

Minus sign indicates that acceleration is in reverse direction

5

2

=

18.75

m

.

Minus sign indicates that acceleration is in reverse direction.

User Igor Skochinsky
by
3.3k points