# A car goes on a straight road 15 m/s, acceleration at a constant rate of speed of 21 m/s^2 in 12 seconds, what is the total distance traveled in 12 seconds?Can you show how you got the answer so I...

A car goes on a straight road 15 m/s, acceleration at a constant rate of speed of 21 m/s^2 in 12 seconds, what is the total distance traveled in 12 seconds?

Can you show how you got the answer so I can understand it. Thanks.

### 5 Answers | Add Yours

Here we have to use the velocity equations.

S = Ut+1/2*a*t^2

If we put this equation from the time that starts acceleration to end of 12 seconds,

S = distance travelled

U = 15m/s

a = 21m/s^2

t = 12s

S = 15*12+1/2*21*12^2

= 1692m

*So in 12s the car has travelled 1692m.*

**Sources:**

The other way to solve this question is as follows:

Let's find velocity after 12 seconds using the relation

v = u + a*t = 15 m/s + 21 m/s^2*12 s = 15 + 252 = 267 m/s

Now we shall use the formula , the total distance travelled in time t is

s = 1/2*(u+v)*t = 1/2*(15+267)*12= 1/2*282*12 = 1692 m

Hence , the distance travelled in 12 s is 1692 m.

we know that

s=u+1/2at^2

therefore

s=15*12+1/2*21*12*12

s=1692m ans.

Here we have to use the velocity equations.

S = Ut+1/2*a*t^2

If we put this equation from the time that starts acceleration to end of 12 seconds,

S = distance travelled

U = 15m/s

a = 21m/s^2

t = 12s

S = 15*12+1/2*21*12^2

= 1692m

Without using kinematics equations, we can use a speed-time graph to visualise the car's speed. At t = 0, the car's speed is 15 m/s. After 12 seconds of accelerating at 21 m/s^2, the car's speed has increased to 21*12+15 = 267 m/s, at t = 12. The total distance travelled by the car is the area under the graph, which is a trapezium. Hence the total distance travelled is 12(15+267)]/2 = **1692 m**.